Question

Evaluate the given indefinite integral: $\int{\dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}dx}$.

Answer 1

Hint: We start solving the problem 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 the necessary calculations to get the required result.

Complete step-by-step solution:
According to the problem, we need to find the given indefinite integral $\int{\dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}dx}$.
We convert the integrand (the 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 polynomials $\left( {{x}^{2}}+4 \right)$ and $\left( {{x}^{2}}+25 \right)$ cannot divide further into 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 denominator. As the polynomials $\left( {{x}^{2}}+4 \right)$ and $\left( {{x}^{2}}+25 \right)$ are not reducible into further polynomials of smaller degrees, we takes numerators as the polynomial with smaller degree. We take the numerators for the partial fractions as $\left( ax+b \right)$ and $\left( cx+d \right)$ for the fractions with denominators $\left( {{x}^{2}}+4 \right)$ and $\left( {{x}^{2}}+25 \right)$.
So, the partial fractions for the given integrand can be written as:
$\Rightarrow \dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}=\dfrac{ax+b}{{{x}^{2}}+4}+\dfrac{cx+d}{{{x}^{2}}+25}$ ---(1).
$\Rightarrow \dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}=\dfrac{\left( \left( ax+b \right).\left( {{x}^{2}}+25 \right) \right)+\left( \left( cx+d \right).\left( {{x}^{2}}+4 \right) \right)}{\left( {{x}^{2}}+4 \right).\left( {{x}^{2}}+25 \right)}$.
$\Rightarrow {{x}^{2}}+1=\left( a{{x}^{3}}+b{{x}^{2}}+25ax+25b \right)+\left( c{{x}^{3}}+d{{x}^{2}}+4cx+4d \right)$.
$\Rightarrow {{x}^{2}}+1=\left( a+c \right){{x}^{3}}+\left( b+d \right){{x}^{2}}+\left( 25a+4c \right)x+\left( 25b+4d \right)$.
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 $b+d=1$.
We get $b=1-d$ ---(3).
We compare coefficients of $x$ on both sides and we get $25a+4c=0$.
From equation (2),
$\Rightarrow 25\left( -c \right)+4c=0$.
$\Rightarrow -25c+4c=0$.
$\Rightarrow -21c=0$.
$\Rightarrow c=0$.
We substitute the value of c in equation (2),
$\Rightarrow a=-0$.
$\Rightarrow a=0$.
We compare constant terms on both sides and we get $25b+4d=1$.
From equation (3),
$\Rightarrow 25\left( 1-d \right)+4d=1$.
$\Rightarrow 25-25d+4d=1$.
$\Rightarrow 25-1=25d-4d$.
$\Rightarrow 24=21d$.
$\Rightarrow d=\dfrac{24}{21}$.
$\Rightarrow d=\dfrac{8}{7}$.
We substitute the value of d in equation (3),
$\Rightarrow b=1-\dfrac{8}{7}$.
$\Rightarrow b=\dfrac{7-8}{7}$.
$\Rightarrow b=\dfrac{-1}{7}$.
Now, we substitute the values of a, b, c, and d in equation (1).
$\Rightarrow \dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}=\dfrac{0x+\left( \dfrac{-1}{7} \right)}{{{x}^{2}}+4}+\dfrac{0x+\left( \dfrac{8}{7} \right)}{{{x}^{2}}+25}$.
$\Rightarrow \dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}=\dfrac{-1}{7\left( {{x}^{2}}+4 \right)}+\dfrac{8}{7\left( {{x}^{2}}+25 \right)}$.
Now, we apply integration on both sides.
$\Rightarrow \int{\dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}dx}=\int{\dfrac{-1}{7\left( {{x}^{2}}+4 \right)}dx}+\int{\dfrac{8}{7\left( {{x}^{2}}+25 \right)}dx}$.
$\Rightarrow \int{\dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}dx}=\dfrac{-1}{7}\times \int{\dfrac{1}{\left( {{x}^{2}}+4 \right)}dx}+\dfrac{8}{7}\times \int{\dfrac{1}{\left( {{x}^{2}}+25 \right)}dx}$.
$\Rightarrow \int{\dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}dx}=\dfrac{-1}{7}\times \int{\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)}dx}+\dfrac{8}{7}\times \int{\dfrac{1}{\left( {{x}^{2}}+{{5}^{2}} \right)}dx}$.
We know that the integration of $\dfrac{1}{\left( {{a}^{2}}+{{x}^{2}} \right)}$ is defined as $\int{\dfrac{1}{\left( {{a}^{2}}+{{x}^{2}} \right)}dx}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)+C$.
$\Rightarrow \int{\dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}dx}=\left( \dfrac{-1}{7}\times \dfrac{1}{2}\times {{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right)+\left( \dfrac{8}{7}\times \dfrac{1}{5}\times {{\tan }^{-1}}\left( \dfrac{x}{5} \right) \right)+C$.
$\Rightarrow \int{\dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}dx}=\left( \dfrac{-{{\tan }^{-1}}\left( \dfrac{x}{2} \right)}{14} \right)+\left( \dfrac{8{{\tan }^{-1}}\left( \dfrac{x}{5} \right)}{35} \right)+C$.
We got the result after performing the integration $\int{\dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}dx}$ as$\left( \dfrac{-{{\tan }^{-1}}\left( \dfrac{x}{2} \right)}{14} \right)+\left( \dfrac{8{{\tan }^{-1}}\left( \dfrac{x}{5} \right)}{35} \right)+C$.
∴ $\int{\dfrac{{{x}^{2}}+1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+25 \right)}dx}=\left( \dfrac{-{{\tan }^{-1}}\left( \dfrac{x}{2} \right)}{14} \right)+\left( \dfrac{8{{\tan }^{-1}}\left( \dfrac{x}{5} \right)}{35} \right)+C$.

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 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 present in the denominator, we always check whether those functions are reducible or not.