How do you use Integration by Substitution to find $\int \frac{x}{x^{4} + 1} d x$ $intx/(x^4+1)dx$ ?

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How do you use Integration by Substitution to find $\int \frac{x}{x^{4} + 1} d x$ $intx/(x^4+1)dx$ ?

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Answer 1 · 2014-09-08T01:38:06

Let $u = x^{2}$ $u=x^2$ .
By taking the derivative,
$\frac{d u}{d x} = 2 x$ ${du}/{dx}=2x$
by taking the reciprocal,
$\Rightarrow \frac{d x}{d u} = \frac{1}{2 x}$ $Rightarrow{dx}/{du}=1/{2x}$
by multiplying by $d u$ $du$ ,
$\Rightarrow d x = \frac{d u}{2 x}$ $Rightarrow dx={du}/{2x}$

By rewriting the integral in terms of $u$ $u$ ,
$\int \frac{x}{x^{4} + 1} d x = \int \frac{x}{u^{2} + 1} \cdot \frac{d u}{2 x}$ $intx/{x^4+1}dx=intx/{u^2+1}cdot{du}/{2x}$
by cancelling out $x$ $x$ 's,
$= \frac{1}{2} \int \frac{1}{1 + u^{2}} d u = \frac{1}{2} {tan}^{- 1} u + C$ $=1/2 int1/{1+u^2}du=1/2tan^{-1}u+C$
by putting $u = x^{2}$ $u=x^2$ back in,
$= \frac{1}{2} {tan}^{- 1} (x^{2}) + C$ $=1/2tan^{-1}(x^2)+C$

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How do you use Integration by Substitution to find $\int \frac{x}{x^{4} + 1} d x$ $intx/(x^4+1)dx$ ?

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