Question

Solve $\int{\dfrac{{{x}^{4}}+1}{{{x}^{6}}+1}}dx$

Answer 1

In this question, to solve the given integral firstly we will add and subtract ${{x}^{2}}$ in the numerator, then split the terms and find the integration of each terms separately by assuming the integrals as ${{I}_{1}}$ and ${{I}_{2}}$ such that $I={{I}_{1}}+{{I}_{2}}$ so that we can find the obtained result.

Complete step by step answer:
Now according to the question we have to find the integral of $\int{\dfrac{{{x}^{4}}+1}{{{x}^{6}}+1}}dx$
Add and subtract ${{x}^{2}}$ in the numerator we get:
$\Rightarrow \int{\dfrac{{{x}^{4}}+1+{{x}^{2}}-{{x}^{2}}}{{{x}^{6}}+1}}dx$
Split the terms:
$\Rightarrow \int{\dfrac{{{x}^{4}}+1-{{x}^{2}}}{{{x}^{6}}+1}}dx+\int{\dfrac{{{x}^{2}}}{{{x}^{6}}+1}dx}$
Let us assume that $\int{\dfrac{{{x}^{4}}+1-{{x}^{2}}}{{{x}^{6}}+1}}dx={{I}_{1}}$ and $\int{\dfrac{{{x}^{2}}}{{{x}^{6}}+1}dx}={{I}_{2}}$ such that $I={{I}_{1}}+{{I}_{2}}$
Where $I=\int{\dfrac{{{x}^{4}}+1-{{x}^{2}}}{{{x}^{6}}+1}}dx+\int{\dfrac{{{x}^{2}}}{{{x}^{6}}+1}dx}$
Integrating ${{I}_{1}}$ we get:
$\Rightarrow {{I}_{1}}=\int{\dfrac{{{x}^{4}}+1-{{x}^{2}}}{{{x}^{6}}+1}}dx$
$\Rightarrow {{I}_{1}}=\int{\dfrac{{{x}^{4}}+1-{{x}^{2}}}{{{({{x}^{2}})}^{3}}+{{1}^{3}}}}dx$
Apply the formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ in the denominator:
$\Rightarrow {{I}_{1}}=\int{\dfrac{{{x}^{4}}+1-{{x}^{2}}}{\left( {{x}^{2}}+1 \right)\left( {{\left( {{x}^{2}} \right)}^{2}}+{{1}^{2}}-{{x}^{2}}\times 1 \right)}}dx$
$\Rightarrow {{I}_{1}}=\int{\dfrac{{{x}^{4}}+1-{{x}^{2}}}{\left( {{x}^{2}}+1 \right)\left( {{x}^{4}}+1-{{x}^{2}} \right)}}dx$
Cancelling out the required terms we get:
$\Rightarrow {{I}_{1}}=\int{\dfrac{1}{\left( {{x}^{2}}+1 \right)}}dx$
$\Rightarrow {{I}_{1}}={{\tan }^{-1}}x+c$
Now integrating ${{I}_{2}}$ we get:
$\Rightarrow {{I}_{2}}=\int{\dfrac{{{x}^{2}}}{{{x}^{6}}+1}dx}$
$\Rightarrow {{I}_{2}}=\int{\dfrac{{{x}^{2}}}{{{\left( {{x}^{3}} \right)}^{2}}+{{1}^{2}}}dx}$
Put ${{x}^{3}}=t$ therefore the differential will be
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{3}} \right)=\dfrac{d}{dx}\left( t \right)$
$\Rightarrow 3{{x}^{2}}=\dfrac{dt}{dx}$
$\Rightarrow {{x}^{2}}dx=\dfrac{dt}{3}$
Put the values we get:
$\Rightarrow {{I}_{2}}=\int{\dfrac{1}{{{\left( t \right)}^{2}}+1}\cdot \dfrac{dt}{3}}$
$\Rightarrow {{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{{{\left( t \right)}^{2}}+1}\cdot dt}$
$\Rightarrow {{I}_{2}}=\dfrac{1}{3}\cdot {{\tan }^{-1}}t+c$
$\Rightarrow {{I}_{2}}=\dfrac{1}{3}\cdot {{\tan }^{-1}}{{\left( x \right)}^{3}}+c$
$\Rightarrow I={{I}_{1}}+{{I}_{2}}$
$\Rightarrow I=\left( {{\tan }^{-1}}x+c \right)+\left( \dfrac{1}{3}\cdot {{\tan }^{-1}}{{\left( x \right)}^{3}}+c \right)$
$\Rightarrow I=\left( {{\tan }^{-1}}x+\dfrac{1}{3}\cdot {{\tan }^{-1}}{{\left( x \right)}^{3}}+2c \right)$
Let $2c=K$ where $K$ is any constant
$\Rightarrow I=\left( {{\tan }^{-1}}x+\dfrac{1}{3}\cdot {{\tan }^{-1}}{{\left( x \right)}^{3}}+K \right)$

Note:
Integrations are essential for calculating the centre of gravity, centre of mass, and predicting the positions of the planets, among other things. Many things employ integration to find their volume, area, and core values. Integration is the reverse process of differentiation.