# Thread: Integration and I dont get along just yet

1. ## Integration and I dont get along just yet

Integrate x^(3/2)ln x dx

Integrate (((x^2)-x+6)/(x^3)+3x) dx

2. $\displaystyle \int x^{\frac{3}{2}}\ln x~dx$

Let $\displaystyle u = \ln x$ and $\displaystyle dv = x^{\frac{3}{2}}~dx$

Then $\displaystyle du = \frac{1}{x}~dx$ and $\displaystyle v = \frac{2}{5}x^{\frac{5}{2}}$

$\displaystyle \int u~dv = u\cdot v - \int v~du$

$\displaystyle \int x^{\frac{3}{2}}\ln x~dx = \frac{2}{5}x^{\frac{5}{2}}\ln x - \int \frac{2}{5}x^{\frac{5}{2}}\frac{1}{x}~dx$

$\displaystyle \frac{2}{5}x^{\frac{5}{2}}\ln x - \frac{2}{5}\int x^{\frac{3}{2}}~dx$

$\displaystyle \frac{2}{5}x^{\frac{5}{2}}\ln x - \frac{2}{5}\cdot\frac{2}{5}x^{\frac{5}{2}}$

$\displaystyle \frac{2}{25}x^{\frac{5}{2}} \cdot \left ( 5\ln x - 2\right )$

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$\displaystyle \int \frac{x^2-x+6}{x^3+3x}~dx$

$\displaystyle \int \frac{x^2+6}{x^3+3x}~dx - \int \frac{x}{x^3+3x}~dx$

Apply partial fraction decomposition on the left integral (That will break its shell). And simplify the right integral.

$\displaystyle \frac{x^2+6}{x^3+3x} = \frac{2}{x}-\frac{x}{x^2+3}$

$\displaystyle \int \frac{2}{x}-\frac{x}{x^2+3}~dx - \int \frac{1}{x^2+3}~dx$

The rest is easy,

$\displaystyle 2\ln x - \frac{1}{2}\ln |x^2+3| - \frac{1}{\sqrt{3}}\arctan \frac{x}{\sqrt{3}}$

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Notes: You can find the integral of $\displaystyle \frac{x}{x^2+3}$ by substituting $\displaystyle u=x^2+3$.

Maybe you wonder how I got that $\displaystyle \frac{1}{\sqrt{3}}\arctan \frac{x}{\sqrt{3}}$. This is how:

$\displaystyle \int \frac{1}{x^2+3}~dx$

$\displaystyle \int \frac{1}{3(\frac{x^2}{3}+1)}~dx$

$\displaystyle \int \frac{1}{3}\cdot\frac{1}{\frac{x^2}{3}+1}~dx$

$\displaystyle \frac{1}{3}\int \frac{1}{\frac{x^2}{3}+1}~dx$

$\displaystyle \frac{1}{3}\int \frac{1}{ { \left ( \frac{x}{\sqrt{3}}\right ) }^2+1}~dx$

Substitute $\displaystyle u=\frac{x}{\sqrt{3}}$

Then, $\displaystyle du = \frac{1}{\sqrt{3}}~dx$

$\displaystyle \frac{1}{3}\int \frac{1}{u^2+1}\sqrt{3}~du$

$\displaystyle \frac{1}{\sqrt{3}}\int \frac{1}{u^2+1}~du$

$\displaystyle \frac{1}{\sqrt{3}}\arctan u$

$\displaystyle \frac{1}{\sqrt{3}}\arctan \frac{x}{\sqrt{3}}$