Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{x^{6}+1}{x^{2}+1} d x$

Solution:

Given:

$\int \frac{x^{6}+1}{x^{2}+1} d x$

By applying: $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$

$\Rightarrow \int \frac{\left(\mathrm{x}^{2}\right)^{3}+(1)^{3}}{\mathrm{x}^{2}+1} \mathrm{dx}$

$\Rightarrow \int \frac{\left(\mathrm{x}^{2}+1\right)\left(\left(\mathrm{x}^{2}\right)^{2}+(1)^{2}-\mathrm{x}^{2} \times 1\right)}{\left(\mathrm{x}^{2}+1\right)} \mathrm{dx}$

$\Rightarrow \int \frac{\left(\mathrm{x}^{2}+1\right)\left(\mathrm{x}^{4}+1-\mathrm{x}^{2}\right)}{\mathrm{x}^{2}+1} \mathrm{dx}$

$\Rightarrow \int\left(\mathrm{x}^{4}+1-\mathrm{x}^{2}\right) \mathrm{dx}$

By Splitting

$\Rightarrow \int x^{4} d x+1 \int d x-\int x^{2} d x$

$\int \mathrm{k} \mathrm{dx}=\mathrm{kx}+\mathrm{c}$

$\Rightarrow \frac{x^{5+1}}{5+1}+x-\frac{x^{3+1}}{3+1}+c$

$\Rightarrow \frac{x^{6}}{6}+x-\frac{x^{4}}{4}+c$

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