How do I evaluate the following integral?

Question

How do I evaluate the following integral?

1 Answer

Impact of this question

1360 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-26T20:24:18

$\frac{1}{2} x^{2} + 18 ln ∣ ∣ x^{2} - 36 ∣ ∣ + C$ $1/2x^2+18ln|x^2-36|+C$

Explanation:

Divide $x^{2} - 36$ $x^2-36$ into $x^{3}$ $x^3$ using polynomial long division.

This yields:

$\int \frac{x^{3}}{x^{2} - 36} d x = \int (x + \frac{36 x}{x^{2} - 36}) d x$ $intx^3/(x^2-36)dx=int(x+(36x)/(x^2-36))dx$

Split this integral up:

$\int x d x + \int \frac{36 x}{x^{2} - 36} d x$ $intxdx + int(36x)/(x^2-36)dx$

$\int x d x = \frac{1}{2} x^{2}$ $intxdx=1/2x^2$

For $\int \frac{36 x}{x^{2} - 36} d x$ $int(36x)/(x^2-36)dx$

$u = x^{2} - 36$ $u=x^2-36$

$d u = 2 x d x$ $du=2xdx$

$18 d u = 36 x d x$ $18du=36xdx$

So, our integral becomes:

$\frac{1}{2} x^{2} + 18 \int \frac{d u}{u} = \frac{1}{2} x^{2} + 18 ln | u | + C = \frac{1}{2} x^{2} + 18 ln ∣ ∣ x^{2} - 36 ∣ ∣ + C$ $1/2x^2+18int(du)/u=1/2x^2+18ln|u|+C=1/2x^2+18ln|x^2-36|+C$

Science

Math

Humanities

... and beyond

How do I evaluate the following integral?

1 Answer

Explanation:

Related questions

Impact of this question