Evaluate the given indefinite integral: $\int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}$.
Answer
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Hint: We start solving this problems by writing the given integrand into the general form of partial fractions. We find the values that were required to convert into partial fractions. After converting into partial fractions, we integrate each of them obtained and make necessary calculations to get the required result.
Complete step-by-step answer:
According to the problem, we need to find the given indefinite integral $\int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}$.
We convert the integrand (function that is to be integrated) into partial fractions, so that we get a denominator that doesn’t contain multiplication of two polynomials.
We can see that the polynomial $\left( {{x}^{2}}+4 \right)$ cannot divide further into the factor of form $\left( x+a \right)\left( a\in R \right)$.
We know that the numerator of the partial fraction will have the degree less than the degree of the denominator. As the polynomials $\left( {{x}^{2}}+4 \right)$ are not reducible into further polynomials of smaller degrees, we take numerators as the polynomial with smaller degree. We take the numerators for the partial fractions as $\left( ax+b \right)$ for the fractions with denominators $\left( {{x}^{2}}+4 \right)$.
We can see that the polynomial ${{x}^{2}}$ is reducible to $x$ and ${{x}^{2}}$. We know that if $R(x)$ of the fraction $\dfrac{Q(x)}{R(x)}$ is decomposed as $R\left( x \right)={{\left( x-p \right)}^{n}}S\left( x \right)$, where $S\left( p \right)\ne 0$, then we get n partial fractions involving $\left( x-p \right)$ as $\dfrac{Q(x)}{R(x)}=\dfrac{{{a}_{1}}}{\left( x-p \right)}+\dfrac{{{a}_{2}}}{{{\left( x-p \right)}^{2}}}+......+\dfrac{{{a}_{n}}}{{{\left( x-p \right)}^{n}}}+\dfrac{{{Q}_{1}}\left( x \right)}{S\left( x \right)}$.
So, the partial fractions for the given integrand can be written as:
\[\Rightarrow \dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}=\dfrac{a}{x}+\dfrac{b}{{{x}^{2}}}+\dfrac{cx+d}{{{x}^{2}}+4}\]---(1).
$\Rightarrow \dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}=\dfrac{\left( a.x.\left( {{x}^{2}}+4 \right) \right)+\left( b.\left( {{x}^{2}}+4 \right) \right)+\left( \left( cx+d \right).{{x}^{2}} \right)}{{{x}^{2}}.\left( {{x}^{2}}+4 \right)}$.
$\Rightarrow 2{{x}^{2}}+4=\left( a{{x}^{3}}+4ax \right)+\left( b{{x}^{2}}+4b \right)+\left( c{{x}^{3}}+d{{x}^{2}} \right)$.
$\Rightarrow 2{{x}^{2}}+4=\left( a+c \right){{x}^{3}}+\left( b+d \right){{x}^{2}}+4ax+4b$.
We compare coefficients of ${{x}^{3}}$ on both sides and we get $a+c=0$.
We get $a=-c$ ---(2).
We compare coefficients of ${{x}^{2}}$ on both sides and we get $b+d=2$.
We get $b=2-d$ ---(3).
We compare coefficients of $x$ on both sides and we get $4a=0$.
We get $a=0$.
We substitute the value of ‘a’ in equation (2),
$\Rightarrow -c=0$.
$\Rightarrow c=0$.
We compare constant terms on both sides and we get $4b=4$.
$\Rightarrow b=\dfrac{4}{4}$.
We get $b=1$.
We substitute the value of ‘b’ in equation (3),
$\Rightarrow 1=2-d$.
$\Rightarrow d=2-1$.
$\Rightarrow d=1$.
Now, we substitute the values of a, b, c and d in equation (1).
\[\Rightarrow \dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}=\dfrac{0}{x}+\dfrac{1}{{{x}^{2}}}+\dfrac{0x+1}{{{x}^{2}}+4}\].
\[\Rightarrow \dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}=\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{2}}+4}\].
Now, we apply integration on both sides.
\[\Rightarrow \int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\int{\dfrac{1}{{{x}^{2}}}dx}+\int{\dfrac{1}{\left( {{x}^{2}}+4 \right)}dx}\].
\[\Rightarrow \int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\int{{{x}^{-2}}dx}+\int{\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)}dx}\].
We know that the integration of $\dfrac{1}{\left( {{a}^{2}}+{{x}^{2}} \right)}$ is defined as $\int{\dfrac{1}{\left( {{a}^{2}}+{{x}^{2}} \right)}dx}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)+C$ and also the integration of ${{x}^{n}}$ is defined as $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$.
\[\Rightarrow \int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\dfrac{{{x}^{-2+1}}}{-2+1}+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C\].
\[\Rightarrow \int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\dfrac{{{x}^{-1}}}{-1}+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C\].
\[\Rightarrow \int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\dfrac{-1}{x}+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C\].
We got the result after performing the integration \[\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}\] as \[\dfrac{-1}{x}+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C\].
∴\[\int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\dfrac{-1}{x}+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C\].
Note:Before converting into fractions, we check whether the degree of the numerator is greater than the degree of the denominator or not. If it is, we divide the numerator until we get the degree of the numerator less than the denominator to convert into partial fractions. If not, we directly convert them into partial fractions. If the polynomials of degree greater than 1 are present in the denominator, we always check whether those functions are reducible or not.
Complete step-by-step answer:
According to the problem, we need to find the given indefinite integral $\int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}$.
We convert the integrand (function that is to be integrated) into partial fractions, so that we get a denominator that doesn’t contain multiplication of two polynomials.
We can see that the polynomial $\left( {{x}^{2}}+4 \right)$ cannot divide further into the factor of form $\left( x+a \right)\left( a\in R \right)$.
We know that the numerator of the partial fraction will have the degree less than the degree of the denominator. As the polynomials $\left( {{x}^{2}}+4 \right)$ are not reducible into further polynomials of smaller degrees, we take numerators as the polynomial with smaller degree. We take the numerators for the partial fractions as $\left( ax+b \right)$ for the fractions with denominators $\left( {{x}^{2}}+4 \right)$.
We can see that the polynomial ${{x}^{2}}$ is reducible to $x$ and ${{x}^{2}}$. We know that if $R(x)$ of the fraction $\dfrac{Q(x)}{R(x)}$ is decomposed as $R\left( x \right)={{\left( x-p \right)}^{n}}S\left( x \right)$, where $S\left( p \right)\ne 0$, then we get n partial fractions involving $\left( x-p \right)$ as $\dfrac{Q(x)}{R(x)}=\dfrac{{{a}_{1}}}{\left( x-p \right)}+\dfrac{{{a}_{2}}}{{{\left( x-p \right)}^{2}}}+......+\dfrac{{{a}_{n}}}{{{\left( x-p \right)}^{n}}}+\dfrac{{{Q}_{1}}\left( x \right)}{S\left( x \right)}$.
So, the partial fractions for the given integrand can be written as:
\[\Rightarrow \dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}=\dfrac{a}{x}+\dfrac{b}{{{x}^{2}}}+\dfrac{cx+d}{{{x}^{2}}+4}\]---(1).
$\Rightarrow \dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}=\dfrac{\left( a.x.\left( {{x}^{2}}+4 \right) \right)+\left( b.\left( {{x}^{2}}+4 \right) \right)+\left( \left( cx+d \right).{{x}^{2}} \right)}{{{x}^{2}}.\left( {{x}^{2}}+4 \right)}$.
$\Rightarrow 2{{x}^{2}}+4=\left( a{{x}^{3}}+4ax \right)+\left( b{{x}^{2}}+4b \right)+\left( c{{x}^{3}}+d{{x}^{2}} \right)$.
$\Rightarrow 2{{x}^{2}}+4=\left( a+c \right){{x}^{3}}+\left( b+d \right){{x}^{2}}+4ax+4b$.
We compare coefficients of ${{x}^{3}}$ on both sides and we get $a+c=0$.
We get $a=-c$ ---(2).
We compare coefficients of ${{x}^{2}}$ on both sides and we get $b+d=2$.
We get $b=2-d$ ---(3).
We compare coefficients of $x$ on both sides and we get $4a=0$.
We get $a=0$.
We substitute the value of ‘a’ in equation (2),
$\Rightarrow -c=0$.
$\Rightarrow c=0$.
We compare constant terms on both sides and we get $4b=4$.
$\Rightarrow b=\dfrac{4}{4}$.
We get $b=1$.
We substitute the value of ‘b’ in equation (3),
$\Rightarrow 1=2-d$.
$\Rightarrow d=2-1$.
$\Rightarrow d=1$.
Now, we substitute the values of a, b, c and d in equation (1).
\[\Rightarrow \dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}=\dfrac{0}{x}+\dfrac{1}{{{x}^{2}}}+\dfrac{0x+1}{{{x}^{2}}+4}\].
\[\Rightarrow \dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}=\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{2}}+4}\].
Now, we apply integration on both sides.
\[\Rightarrow \int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\int{\dfrac{1}{{{x}^{2}}}dx}+\int{\dfrac{1}{\left( {{x}^{2}}+4 \right)}dx}\].
\[\Rightarrow \int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\int{{{x}^{-2}}dx}+\int{\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)}dx}\].
We know that the integration of $\dfrac{1}{\left( {{a}^{2}}+{{x}^{2}} \right)}$ is defined as $\int{\dfrac{1}{\left( {{a}^{2}}+{{x}^{2}} \right)}dx}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)+C$ and also the integration of ${{x}^{n}}$ is defined as $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$.
\[\Rightarrow \int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\dfrac{{{x}^{-2+1}}}{-2+1}+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C\].
\[\Rightarrow \int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\dfrac{{{x}^{-1}}}{-1}+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C\].
\[\Rightarrow \int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\dfrac{-1}{x}+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C\].
We got the result after performing the integration \[\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}\] as \[\dfrac{-1}{x}+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C\].
∴\[\int{\dfrac{2{{x}^{2}}+4}{{{x}^{2}}\left( {{x}^{2}}+4 \right)}dx}=\dfrac{-1}{x}+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C\].
Note:Before converting into fractions, we check whether the degree of the numerator is greater than the degree of the denominator or not. If it is, we divide the numerator until we get the degree of the numerator less than the denominator to convert into partial fractions. If not, we directly convert them into partial fractions. If the polynomials of degree greater than 1 are present in the denominator, we always check whether those functions are reducible or not.
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