Question

Solve the integral function: $\int{\dfrac{1}{3{{x}^{2}}+13x+10}dx}$

Answer 1

Hint: First convert $\dfrac{1}{3{{x}^{2}}+13x+10}$ in the form of $\dfrac{E}{\left( 3x+10 \right)}+\dfrac{F}{\left( x+1 \right)}$, then apply the formula of $\int{\dfrac{dx}{Ax+B}}$ which is equal to $\dfrac{1}{\log }\left( Ax+B \right)+C.$ Then add them together to get the answer.

Complete step-by-step answer:
The given integral is,
$\int{\dfrac{dx}{3{{x}^{2}}+13x+10}}$
Now we will factorize the denominator by using the middle term factor. So, $3{{x}^{2}}+13x+10$ can be written as $3{{x}^{2}}+3x+10x+10$ which can further factorized and written as (3x + 10)(x + 1).
So, the given integral can be written as,
$\int{\dfrac{dx}{3{{x}^{2}}+13x+10}}=\int{\dfrac{dx}{\left( 3x+10 \right)\left( x+1 \right)}}$
If the integration is in form of $\int{\dfrac{dx}{\left( Ax+B \right)\left( Cx+D \right)}}$ then we can write as sum of $\int{\dfrac{Edx}{Ax+B}}$ and \[\int{\dfrac{Fdx}{Cx+D}}\]
Here A, B, C, D, E, F are constants.
So, at first we will express the fraction $\dfrac{1}{\left( 3x+10 \right)\left( x+1 \right)}$ as $\dfrac{E}{\left( 3x+10 \right)}+\dfrac{F}{\left( x+1 \right)}$, where E and F are constants. So we can write,
$\dfrac{1}{\left( 3x+10 \right)\left( x+1 \right)}=\dfrac{E}{3x+10}+\dfrac{F}{x+1}.........(i)$
Hence by cross multiplication, the numerator can be written as,
1 = E (x + 1) + F (3x + 10)
So, after separating ‘x’ terms and constant term we can write it as,
1 = x (E + 3F) + (E + 10F)
Now comparing the coefficients we get,
E + 3F = 0 as in the left hand side there are no terms related to x.
E + 10F = 1 as in the right hand side 1 is the constant term.
Now as we know,
E + 3F = 0
So we can write it as,
E = -3F
Now substituting, E = -3F in E + 10F = 1 we get,
-3F + 10F = 1
So,
$F=\dfrac{1}{7}$
Now as we know, E = -3F, so we get
$E=\dfrac{-3}{7}$
Hence we can write the equation (i) as,
$\dfrac{1}{\left( 3x+10 \right)\left( x+1 \right)}=\dfrac{-3}{7}\left\{ \dfrac{1}{\left( 3x+10 \right)} \right\}+\dfrac{1}{7}\left\{ \dfrac{1}{\left( x+1 \right)} \right\}$
Now we will integrate,
$\int{\dfrac{dx}{\left( 3x+10 \right)\left( x+1 \right)}=\dfrac{-3}{7}\int{\dfrac{dx}{3x+10}+\dfrac{1}{7}\int{\dfrac{dx}{x+1}}}}$
Here we will apply formula that $\int{\dfrac{dx}{Ax+B}=\dfrac{\log \left( Ax+B \right)}{A}+C}$ where A, B, C are the constant terms.
So we can write $\int{\dfrac{dx}{3x+10}=\dfrac{\log \left( 3x+10 \right)}{3}}+C$ where C be any constant term and $\int{\dfrac{dx}{x+2}=\log x+1+{{C}_{2}}}$ where ${{C}_{2}}$ is also a constant term
So now we can write,
$\int{\dfrac{dx}{\left( 3x+10 \right)\left( x+1 \right)}=\dfrac{-3}{7}\dfrac{\log \left( 3x+10 \right)}{3}+\dfrac{1}{7}\log \left( x+1 \right)+{{C}_{1}}+{{C}_{2}}}$
Which is simplified as,
\[\int{\dfrac{dx}{\left( 3x+10 \right)\left( x+1 \right)}=}-\dfrac{\log \left( 3x+10 \right)}{7}+\dfrac{\log \left( x+1 \right)}{7}+C\]
Here ‘C’ is the sum of the constants.
The answer of integration is $-\dfrac{\log \left( 3x+10 \right)}{7}+\dfrac{\log \left( x+1 \right)}{7}+C$

Note: Students must be careful while taking middle term factors and changing them into products of linear factors. They should also be careful about the calculation while taking out the value of coefficients.
Another approach is to convert the denominator in the form of $\left( {{x}^{2}}-{{a}^{2}} \right)$, the apply the formula, $\int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{1}{2a}\log \left| \dfrac{x-a}{x+a} \right|+C}$
In this method also you will get the same answer.