# Evaluating an integral

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• May 18th 2011, 12:41 AM
brumby_3
Evaluating an integral
Evaluate the integral ∫ [x^3/sqrt(x^2+1)] in two different ways.

a) Integration by parts
b) The substitution u = sqrt(x^2+1)

Not sure how to do this, help would be greatly appreciated!
• May 18th 2011, 01:36 AM
chisigma
Quote:

Originally Posted by brumby_3
Evaluate the integral ∫ [x^3/sqrt(x^2+1)] in two different ways.

a) Integration by parts
b) The substitution u = sqrt(x^2+1)

Not sure how to do this, help would be greatly appreciated!

a) by parts...

... taking into account the identity...

http://quicklatex.com/cache3/ql_f8ed...4a24645_l3.png (1)

... is...

http://quicklatex.com/cache3/ql_0e65...8fda6c7_l3.png

http://quicklatex.com/cache3/ql_37da...292da11_l3.png (2)

Kind regards

$\chi$ $\sigma$
• May 18th 2011, 01:55 AM
TheCoffeeMachine
Quote:

Originally Posted by brumby_3

b) The substitution u = sqrt(x^2+1)

\begin{aligned}& u: = \sqrt{x^2+1} \Rightarrow \frac{du}{dx} = \frac{2x}{2\sqrt{x^2+1}} = \frac{x}{\sqrt{x^2+1}} \Rightarrow dx = \frac{\sqrt{x^2+1}}{x}\;{du} \\& \therefore \int\frac{x^3}{\sqrt{x^2+1}}\;{dx} = \int\frac{x^3\sqrt{x^2+1}}{x\sqrt{x^2+1}}\;{du} = \int x^2\;{du} = \int (u^2-1)\;{du} = \cdots \end{aligned}