How do you evaluate the integral $ \int{\dfrac{dx}{{{x}^{3}}+1}} $ ?
Answer
Verified495.3k+ views
Hint: In this question, we need to find the integral of $ \dfrac{1}{1+{{x}^{3}}} $ with respect to x. For this, we will first factorize $ 1+{{x}^{3}} $ using the identity $ \left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) $ . Then we will use a partial fraction to change the form $ \dfrac{1}{1+{{x}^{3}}} $ as the sum of terms which can be integrated easily. After that we will apply integration on each term and evaluate the integral.
Complete step by step answer:
Here we need to evaluate the integral $ \int{\dfrac{dx}{{{x}^{3}}+1}} $ . For this, let us first factorize the denominator of the given function $ \dfrac{1}{1+{{x}^{3}}} $ .
The denominator is $ {{x}^{3}}+1 $ . We know that $ {{\left( 1 \right)}^{3}}=1 $ . So we can say $ {{x}^{3}}+1={{x}^{3}}+{{1}^{3}} $ . It is of the form $ \left( {{a}^{3}}+{{b}^{3}} \right) $ so we can use the identity $ \left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) $ . We get $ \left( {{x}^{3}}+{{1}^{3}} \right)=\left( x+1 \right)\left( {{x}^{2}}-x+{{1}^{2}} \right) $ .
Thus the function becomes $ \dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)} $ .
Now let us use partial fraction for this i.e. we need to change the term $ \dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)} $ into sum of some terms that can be integral easily.
$ \dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)} $ will be changed in the form as $ \dfrac{Ax+B}{\left( {{x}^{2}}-x+1 \right)}+\dfrac{C}{\left( x+1 \right)} $ so we have $ \dfrac{1}{{{x}^{3}}+1}=\dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)}=\dfrac{Ax+B}{{{x}^{2}}-x+1}+\dfrac{C}{x+1} $ .
Taking LCM as $ \left( x+1 \right)\left( {{x}^{2}}-x+1 \right) $ in the right side we get $ \dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)}=\dfrac{\left( Ax+B \right)\left( x+1 \right)+C\left( {{x}^{2}}-x+1 \right)}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)} $ .
Thus we will solve the following equation to find the value of A, B and C.
$ 1=\left( Ax+B \right)\left( x+1 \right)+C\left( {{x}^{2}}-x+1 \right) $ .
Let x = -1 we get $ 1=\left( -A+B \right)\left( -1+1 \right)+C\left( 1+1+1 \right)\Rightarrow 1=3C $ .
Simplifying it we get $ C=\dfrac{1}{3} $ .
Now putting value of C we get $ 1=\left( Ax+B \right)\left( x+1 \right)+\dfrac{1}{3}\left( {{x}^{2}}-x+1 \right) $ .
Taking x = 0 we get $ 1=\left( A\left( 0 \right)+B \right)\left( 0+1 \right)+\dfrac{1}{3}\left( 0-0+1 \right)\Rightarrow 1=B+\dfrac{1}{3} $ .
Simplifying it we get $ B=1-\dfrac{1}{3} $ .
Simplifying it we get $ B=\dfrac{2}{3} $ .
Now let x = 1 we get $ 1=\left( A+\dfrac{2}{3} \right)\left( 2 \right)+\dfrac{1}{3} $ .
Simplifying it we get $ 1=2A+\dfrac{5}{3} $ .
Simplifying it we get $ A=\dfrac{-1}{3} $ .
Hence we have found the value of A, B and C as \[\dfrac{-1}{3},\dfrac{2}{3}\text{ and }\dfrac{1}{3}\] respectively.
So the function becomes \[\dfrac{1}{{{x}^{3}}+1}=\dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)}=\dfrac{\dfrac{-1}{3}x+\dfrac{2}{3}}{{{x}^{2}}-x+1}+\dfrac{\dfrac{1}{3}}{x+1}\].
Simplifying it we get \[\dfrac{1}{{{x}^{3}}+1}=\dfrac{-1}{3}\left( \dfrac{x-2}{{{x}^{2}}-x+1} \right)+\dfrac{1}{3}\left( \dfrac{1}{x+1} \right)\].
Taking integral we get \[\int{\dfrac{1}{{{x}^{3}}+1}dx}=\int{\dfrac{-1}{3}\left( \dfrac{x-2}{{{x}^{2}}-x+1} \right)}dx+\int{\dfrac{1}{3}\left( \dfrac{1}{x+1} \right)}dx\].
Let us solve the integrals separately.
Taking $ {{I}_{1}}=\dfrac{-1}{3}\int{\dfrac{x-2}{{{x}^{2}}-x+1}dx}\text{ and }{{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{x+1}dx} $ .
So we get $ \int{\dfrac{1}{1+{{x}^{3}}}dx}={{I}_{1}}+{{I}_{2}}\cdots \cdots \cdots \left( 1 \right) $ .
Solving $ {{I}_{2}} $ first $ {{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{x+1}dx} $ .
Let us substitute x+1 as u we get x+1 = u.
Taking derivative on both sides we get 1 = du so we get $ {{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{u}dx} $ .
As we know that, $ \int{\dfrac{1}{x}dx}=\ln x+c $ so we get $ {{I}_{2}}=\dfrac{1}{3}\ln u+c $ .
Substituting value of u as (x+1) we get $ {{I}_{2}}=\dfrac{1}{3}\ln \left( x+1 \right)+c\cdots \cdots \cdots \left( 2 \right) $ .
Solving $ {{I}_{1}} $ , $ {{I}_{1}}=\dfrac{-1}{3}\int{\dfrac{x-2}{{{x}^{2}}-x+1}dx} $ .
To make the numerator as derivative of denominator let us multiply and divide function by 2 we get $ {{I}_{1}}=\dfrac{-1}{6}\int{\dfrac{2x-4}{{{x}^{2}}-x+1}dx} $ .
We still want the numerator as 2x-1. So let us split 4 into 1+3 we get $ {{I}_{1}}=\dfrac{-1}{6}\int{\dfrac{\left( 2x-1-3 \right)}{{{x}^{2}}-x+1}dx} $ .
Now let us separate the denominator for two numerator (2x-1) and -3 we get $ {{I}_{1}}=\dfrac{-1}{6}\int{\left( \dfrac{2x-1}{{{x}^{2}}-x+1}-\dfrac{3}{{{x}^{2}}-x+1} \right)dx} $ .
Separating the integrals we get $ {{I}_{1}}=\dfrac{-1}{6}\int{\left( \dfrac{2x-1}{{{x}^{2}}-x+1}dx+\dfrac{3}{6} \right)\dfrac{1}{{{x}^{2}}-x+1}dx} $ .
Simplifying it we get $ {{I}_{1}}=\dfrac{-1}{6}\int{\left( \dfrac{2x-1}{{{x}^{2}}-x+1}dx+\dfrac{1}{2} \right)\dfrac{1}{{{x}^{2}}-x+1}dx} $ .
Now let us solve them separately again.
Taking \[{{I}_{1}}'=\dfrac{-1}{6}\int{\dfrac{2x-1}{{{x}^{2}}-x+1}dx}\text{ and }{{I}_{1}}''=\dfrac{1}{2}\int{\dfrac{1}{{{x}^{2}}-x+1}dx}\].
So $ {{I}_{1}} $ becomes $ {{I}_{1}}={{I}_{1}}'+{{I}_{1}}''\cdots \cdots \cdots \left( 3 \right) $ .
Solving $ {{I}_{1}}' $ we get \[{{I}_{1}}'=\dfrac{-1}{6}\int{\dfrac{2x-1}{{{x}^{2}}-x+1}dx}\].
Let us substitute \[{{x}^{2}}-x+1\] as u we get \[{{x}^{2}}-x+1=u\].
Taking derivative on both sides we get $ \left( 2x-1 \right)dx=du $ .
So the integral becomes \[{{I}_{1}}'=\dfrac{-1}{6}\int{\dfrac{1}{u}dx}\].
We know that $ \int{\dfrac{1}{x}dx}=\ln x $ so we get \[{{I}_{1}}'=\dfrac{-1}{6}\ln u\].
Substituting back the value of u as \[{{x}^{2}}-x+1\] we get \[{{I}_{1}}'=\dfrac{-1}{6}\ln \left( {{x}^{2}}-x+1 \right)\cdots \cdots \cdots \left( 4 \right)\].
Solving \[{{I}_{1}}''\] we have \[{{I}_{1}}''=\dfrac{1}{2}\int{\dfrac{1}{{{x}^{2}}-x+1}dx}\].
We have the denominator as \[{{x}^{2}}-x+1\]. Let us use the square method here. Adding and subtracting $ \dfrac{1}{4} $ in the expression we get \[{{x}^{2}}-x+1+\dfrac{1}{4}-\dfrac{1}{4}\].
Combining \[{{x}^{2}}-x+\dfrac{1}{4}\] as $ {{\left( x-\dfrac{1}{2} \right)}^{2}} $ and solving $ 1-\dfrac{1}{4} $ as $ \dfrac{3}{4} $ we get \[{{x}^{2}}-x+1={{\left( x-\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}\].
Taking LCM of 2 in the whole square term we get \[{{x}^{2}}-x+1={{\left( \dfrac{2x-1}{2} \right)}^{2}}+\dfrac{3}{4}\].
Simplifying we get \[{{x}^{2}}-x+1=\dfrac{{{\left( 2x-1 \right)}^{2}}}{4}+\dfrac{3}{4}\].
Therefore \[{{x}^{2}}-x+1=\dfrac{1}{4}\left( {{\left( 2x-1 \right)}^{2}}+3 \right)\].
So \[{{I}_{1}}''\] becomes \[{{I}_{1}}''=\dfrac{1}{2}\int{\dfrac{1}{\dfrac{1}{4}\left( {{\left( 2x-1 \right)}^{2}}+3 \right)}dx}\].
Solving for \[\dfrac{1}{2}\text{ and }\dfrac{1}{4}\] we get \[{{I}_{1}}''=2\int{\dfrac{1}{{{\left( 2x-1 \right)}^{2}}+3}dx}\].
We can write 3 as $ {{\left( \sqrt{3} \right)}^{2}} $ so we get \[{{I}_{1}}''=2\int{\dfrac{1}{{{\left( 2x-1 \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}}dx}\].
We know that $ \int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}}\left( \dfrac{x}{a} \right) $ .
For x as (2x-1) and a as $ \sqrt{3} $ we will have \[\int{\dfrac{1}{{{\left( 2x-1 \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}}}=\dfrac{1}{2\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x-1}{\sqrt{3}} \right)\].
(We have taken 2 as the denominator because the coefficient of x is 2 in squared terms).
So the integral becomes \[{{I}_{1}}''=\dfrac{2}{2\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x-1}{\sqrt{3}} \right)\].
Simplifying we get \[{{I}_{1}}''=\dfrac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x-1}{\sqrt{3}} \right)\cdots \cdots \cdots \left( 5 \right)\].
Putting values of (4) and (5) in (3) we get \[{{I}_{1}}=\dfrac{-1}{6}\ln \left( {{x}^{2}}-x+1 \right)+\dfrac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x-1}{\sqrt{3}} \right)\cdots \cdots \cdots \left( 6 \right)\].
Putting values from (6) and (2) in (1) we get \[\int{\dfrac{1}{1+{{x}^{3}}}}dx=\dfrac{1}{3}\ln \left( x+1 \right)-\dfrac{1}{6}\ln \left( {{x}^{2}}-x+1 \right)+\dfrac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x-1}{\sqrt{3}} \right)+c\].
Which is our required integral.
Note:
Students should keep in mind the formula of integral and partial fractions to solve this sum. Note that in $ \int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}}dx $ if coefficient of x is b then we divide the answer by b. Take care of signs and constant terms. Note that for indefinite integral, adding a and c at the end is necessary.
Complete step by step answer:
Here we need to evaluate the integral $ \int{\dfrac{dx}{{{x}^{3}}+1}} $ . For this, let us first factorize the denominator of the given function $ \dfrac{1}{1+{{x}^{3}}} $ .
The denominator is $ {{x}^{3}}+1 $ . We know that $ {{\left( 1 \right)}^{3}}=1 $ . So we can say $ {{x}^{3}}+1={{x}^{3}}+{{1}^{3}} $ . It is of the form $ \left( {{a}^{3}}+{{b}^{3}} \right) $ so we can use the identity $ \left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) $ . We get $ \left( {{x}^{3}}+{{1}^{3}} \right)=\left( x+1 \right)\left( {{x}^{2}}-x+{{1}^{2}} \right) $ .
Thus the function becomes $ \dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)} $ .
Now let us use partial fraction for this i.e. we need to change the term $ \dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)} $ into sum of some terms that can be integral easily.
$ \dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)} $ will be changed in the form as $ \dfrac{Ax+B}{\left( {{x}^{2}}-x+1 \right)}+\dfrac{C}{\left( x+1 \right)} $ so we have $ \dfrac{1}{{{x}^{3}}+1}=\dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)}=\dfrac{Ax+B}{{{x}^{2}}-x+1}+\dfrac{C}{x+1} $ .
Taking LCM as $ \left( x+1 \right)\left( {{x}^{2}}-x+1 \right) $ in the right side we get $ \dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)}=\dfrac{\left( Ax+B \right)\left( x+1 \right)+C\left( {{x}^{2}}-x+1 \right)}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)} $ .
Thus we will solve the following equation to find the value of A, B and C.
$ 1=\left( Ax+B \right)\left( x+1 \right)+C\left( {{x}^{2}}-x+1 \right) $ .
Let x = -1 we get $ 1=\left( -A+B \right)\left( -1+1 \right)+C\left( 1+1+1 \right)\Rightarrow 1=3C $ .
Simplifying it we get $ C=\dfrac{1}{3} $ .
Now putting value of C we get $ 1=\left( Ax+B \right)\left( x+1 \right)+\dfrac{1}{3}\left( {{x}^{2}}-x+1 \right) $ .
Taking x = 0 we get $ 1=\left( A\left( 0 \right)+B \right)\left( 0+1 \right)+\dfrac{1}{3}\left( 0-0+1 \right)\Rightarrow 1=B+\dfrac{1}{3} $ .
Simplifying it we get $ B=1-\dfrac{1}{3} $ .
Simplifying it we get $ B=\dfrac{2}{3} $ .
Now let x = 1 we get $ 1=\left( A+\dfrac{2}{3} \right)\left( 2 \right)+\dfrac{1}{3} $ .
Simplifying it we get $ 1=2A+\dfrac{5}{3} $ .
Simplifying it we get $ A=\dfrac{-1}{3} $ .
Hence we have found the value of A, B and C as \[\dfrac{-1}{3},\dfrac{2}{3}\text{ and }\dfrac{1}{3}\] respectively.
So the function becomes \[\dfrac{1}{{{x}^{3}}+1}=\dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)}=\dfrac{\dfrac{-1}{3}x+\dfrac{2}{3}}{{{x}^{2}}-x+1}+\dfrac{\dfrac{1}{3}}{x+1}\].
Simplifying it we get \[\dfrac{1}{{{x}^{3}}+1}=\dfrac{-1}{3}\left( \dfrac{x-2}{{{x}^{2}}-x+1} \right)+\dfrac{1}{3}\left( \dfrac{1}{x+1} \right)\].
Taking integral we get \[\int{\dfrac{1}{{{x}^{3}}+1}dx}=\int{\dfrac{-1}{3}\left( \dfrac{x-2}{{{x}^{2}}-x+1} \right)}dx+\int{\dfrac{1}{3}\left( \dfrac{1}{x+1} \right)}dx\].
Let us solve the integrals separately.
Taking $ {{I}_{1}}=\dfrac{-1}{3}\int{\dfrac{x-2}{{{x}^{2}}-x+1}dx}\text{ and }{{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{x+1}dx} $ .
So we get $ \int{\dfrac{1}{1+{{x}^{3}}}dx}={{I}_{1}}+{{I}_{2}}\cdots \cdots \cdots \left( 1 \right) $ .
Solving $ {{I}_{2}} $ first $ {{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{x+1}dx} $ .
Let us substitute x+1 as u we get x+1 = u.
Taking derivative on both sides we get 1 = du so we get $ {{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{u}dx} $ .
As we know that, $ \int{\dfrac{1}{x}dx}=\ln x+c $ so we get $ {{I}_{2}}=\dfrac{1}{3}\ln u+c $ .
Substituting value of u as (x+1) we get $ {{I}_{2}}=\dfrac{1}{3}\ln \left( x+1 \right)+c\cdots \cdots \cdots \left( 2 \right) $ .
Solving $ {{I}_{1}} $ , $ {{I}_{1}}=\dfrac{-1}{3}\int{\dfrac{x-2}{{{x}^{2}}-x+1}dx} $ .
To make the numerator as derivative of denominator let us multiply and divide function by 2 we get $ {{I}_{1}}=\dfrac{-1}{6}\int{\dfrac{2x-4}{{{x}^{2}}-x+1}dx} $ .
We still want the numerator as 2x-1. So let us split 4 into 1+3 we get $ {{I}_{1}}=\dfrac{-1}{6}\int{\dfrac{\left( 2x-1-3 \right)}{{{x}^{2}}-x+1}dx} $ .
Now let us separate the denominator for two numerator (2x-1) and -3 we get $ {{I}_{1}}=\dfrac{-1}{6}\int{\left( \dfrac{2x-1}{{{x}^{2}}-x+1}-\dfrac{3}{{{x}^{2}}-x+1} \right)dx} $ .
Separating the integrals we get $ {{I}_{1}}=\dfrac{-1}{6}\int{\left( \dfrac{2x-1}{{{x}^{2}}-x+1}dx+\dfrac{3}{6} \right)\dfrac{1}{{{x}^{2}}-x+1}dx} $ .
Simplifying it we get $ {{I}_{1}}=\dfrac{-1}{6}\int{\left( \dfrac{2x-1}{{{x}^{2}}-x+1}dx+\dfrac{1}{2} \right)\dfrac{1}{{{x}^{2}}-x+1}dx} $ .
Now let us solve them separately again.
Taking \[{{I}_{1}}'=\dfrac{-1}{6}\int{\dfrac{2x-1}{{{x}^{2}}-x+1}dx}\text{ and }{{I}_{1}}''=\dfrac{1}{2}\int{\dfrac{1}{{{x}^{2}}-x+1}dx}\].
So $ {{I}_{1}} $ becomes $ {{I}_{1}}={{I}_{1}}'+{{I}_{1}}''\cdots \cdots \cdots \left( 3 \right) $ .
Solving $ {{I}_{1}}' $ we get \[{{I}_{1}}'=\dfrac{-1}{6}\int{\dfrac{2x-1}{{{x}^{2}}-x+1}dx}\].
Let us substitute \[{{x}^{2}}-x+1\] as u we get \[{{x}^{2}}-x+1=u\].
Taking derivative on both sides we get $ \left( 2x-1 \right)dx=du $ .
So the integral becomes \[{{I}_{1}}'=\dfrac{-1}{6}\int{\dfrac{1}{u}dx}\].
We know that $ \int{\dfrac{1}{x}dx}=\ln x $ so we get \[{{I}_{1}}'=\dfrac{-1}{6}\ln u\].
Substituting back the value of u as \[{{x}^{2}}-x+1\] we get \[{{I}_{1}}'=\dfrac{-1}{6}\ln \left( {{x}^{2}}-x+1 \right)\cdots \cdots \cdots \left( 4 \right)\].
Solving \[{{I}_{1}}''\] we have \[{{I}_{1}}''=\dfrac{1}{2}\int{\dfrac{1}{{{x}^{2}}-x+1}dx}\].
We have the denominator as \[{{x}^{2}}-x+1\]. Let us use the square method here. Adding and subtracting $ \dfrac{1}{4} $ in the expression we get \[{{x}^{2}}-x+1+\dfrac{1}{4}-\dfrac{1}{4}\].
Combining \[{{x}^{2}}-x+\dfrac{1}{4}\] as $ {{\left( x-\dfrac{1}{2} \right)}^{2}} $ and solving $ 1-\dfrac{1}{4} $ as $ \dfrac{3}{4} $ we get \[{{x}^{2}}-x+1={{\left( x-\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}\].
Taking LCM of 2 in the whole square term we get \[{{x}^{2}}-x+1={{\left( \dfrac{2x-1}{2} \right)}^{2}}+\dfrac{3}{4}\].
Simplifying we get \[{{x}^{2}}-x+1=\dfrac{{{\left( 2x-1 \right)}^{2}}}{4}+\dfrac{3}{4}\].
Therefore \[{{x}^{2}}-x+1=\dfrac{1}{4}\left( {{\left( 2x-1 \right)}^{2}}+3 \right)\].
So \[{{I}_{1}}''\] becomes \[{{I}_{1}}''=\dfrac{1}{2}\int{\dfrac{1}{\dfrac{1}{4}\left( {{\left( 2x-1 \right)}^{2}}+3 \right)}dx}\].
Solving for \[\dfrac{1}{2}\text{ and }\dfrac{1}{4}\] we get \[{{I}_{1}}''=2\int{\dfrac{1}{{{\left( 2x-1 \right)}^{2}}+3}dx}\].
We can write 3 as $ {{\left( \sqrt{3} \right)}^{2}} $ so we get \[{{I}_{1}}''=2\int{\dfrac{1}{{{\left( 2x-1 \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}}dx}\].
We know that $ \int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}}\left( \dfrac{x}{a} \right) $ .
For x as (2x-1) and a as $ \sqrt{3} $ we will have \[\int{\dfrac{1}{{{\left( 2x-1 \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}}}=\dfrac{1}{2\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x-1}{\sqrt{3}} \right)\].
(We have taken 2 as the denominator because the coefficient of x is 2 in squared terms).
So the integral becomes \[{{I}_{1}}''=\dfrac{2}{2\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x-1}{\sqrt{3}} \right)\].
Simplifying we get \[{{I}_{1}}''=\dfrac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x-1}{\sqrt{3}} \right)\cdots \cdots \cdots \left( 5 \right)\].
Putting values of (4) and (5) in (3) we get \[{{I}_{1}}=\dfrac{-1}{6}\ln \left( {{x}^{2}}-x+1 \right)+\dfrac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x-1}{\sqrt{3}} \right)\cdots \cdots \cdots \left( 6 \right)\].
Putting values from (6) and (2) in (1) we get \[\int{\dfrac{1}{1+{{x}^{3}}}}dx=\dfrac{1}{3}\ln \left( x+1 \right)-\dfrac{1}{6}\ln \left( {{x}^{2}}-x+1 \right)+\dfrac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x-1}{\sqrt{3}} \right)+c\].
Which is our required integral.
Note:
Students should keep in mind the formula of integral and partial fractions to solve this sum. Note that in $ \int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}}dx $ if coefficient of x is b then we divide the answer by b. Take care of signs and constant terms. Note that for indefinite integral, adding a and c at the end is necessary.
Recently Updated Pages
Master Class 10 English: Engaging Questions & Answers for Success
Master Class 10 Social Science: Engaging Questions & Answers for Success
Master Class 10 Maths: Engaging Questions & Answers for Success
Master Class 10 Science: Engaging Questions & Answers for Success
Class 10 Question and Answer - Your Ultimate Solutions Guide
Master Class 10 Computer Science: Engaging Questions & Answers for Success
Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE
When and how did Canada eventually gain its independence class 10 social science CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Write examples of herbivores carnivores and omnivo class 10 biology CBSE
Which planet is known as the Watery Planet AJupiter class 10 social science CBSE