Integral of x/((x^2+9)^0.5)

\int\frac{x}{\sqrt{x^2+9}}dx

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Answer

$\sqrt{9+x^2}+C_0$

Step by step solution

Problem

$\int\frac{x}{\sqrt{x^2+9}}dx$
1

Solve the integral $\int\frac{x}{\sqrt{9+x^2}}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=9+x^2 \\ du=2xdx\end{matrix}$
2

Isolate $dx$ in the previous equation

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

Substituting $u$ and $dx$ in the integral

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

Taking the constant out of the integral

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

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\frac{1}{2}\int u^{-\frac{1}{2}}du$
6

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{1}{2}\cdot 2\sqrt{u}$
7

Substitute $u$ back for it's value, $9+x^2$

$1\sqrt{9+x^2}$
8

Any expression multiplied by $1$ is equal to itself

$\sqrt{9+x^2}$
9

Add the constant of integration

$\sqrt{9+x^2}+C_0$

Answer

$\sqrt{9+x^2}+C_0$

Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.5 seconds

Views:

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