Integral of x/(4-1x^2)

\int\frac{x}{4-x^2}dx

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Answer

$-\frac{1}{2}\ln\left|4-x^2\right|+C_0$

Step by step solution

Problem

$\int\frac{x}{4-x^2}dx$
1

Solve the integral $\int\frac{x}{4-x^2}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=4-x^2 \\ du=-2xdx\end{matrix}$
2

Isolate $dx$ in the previous equation

$\frac{du}{-2x}=dx$
3

Substituting $u$ and $dx$ in the integral

$\int\frac{1}{-2u}du$
4

Taking the constant out of the integral

$-\frac{1}{2}\int\frac{1}{u}du$
5

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$-\frac{1}{2}\cdot 1\ln\left|u\right|$
6

Substitute $u$ back for it's value, $4-x^2$

$-\frac{1}{2}\ln\left|4-x^2\right|$
7

Add the constant of integration

$-\frac{1}{2}\ln\left|4-x^2\right|+C_0$

Answer

$-\frac{1}{2}\ln\left|4-x^2\right|+C_0$

Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.32 seconds

Views:

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