Substitution questions

**nirva** · November 13th 2006, 07:50 PM

These are some questions on a practice quiz that I can't get, a little help please.

$\int \frac{b}{x^2+a^2} dx$

$\int \sec(3x)\tan(3x)dx$

$\int \frac{2x+13}{x^2+6x+15} dx$

$\int \frac{8}{x^2-9} dx$

Also this one:
"Fill in for the start of a Partial Fraction Decomposition - do not actually solve"
$\frac{some-quadratic}{x^3+x^2+3x+3}$

Thanks a lot

**ThePerfectHacker** · November 13th 2006, 07:56 PM

Originally Posted by nirva

These are some questions on a practice quiz that I can't get, a little help please.

$\int \frac{b}{x^2+a^2} dx$

If $a\not = 0$ Divide by it,
$\frac{b}{a^2}\int \frac{1}{1+x^2/a^2}dx$
For simplicify sake say $a>0$
Then,
$\frac{b}{a^2}\int \frac{1}{1+(x/a)^2}dx$
The substitution function $u=x/a$ will fast transform this into an inverse tangent.
Thus,
$\frac{b}{a}\tan^{-1}(x/a)+C$

$\int \frac{8}{x^2-9} dx$

Fractional decomposition,
$\frac{8}{6}\int \frac{1}{x-3}-\frac{1}{x+3}dx$
Thus,
$\frac{4}{3}\ln |x-3|-\frac{4}{3}\ln |x+3|+C=\frac{4}{3}\ln \left| \frac{x-3}{x+3} \right|+C$

**Soroban** · November 13th 2006, 08:10 PM

Hello, nirva!

Fill in for the start of a Partial Fraction Decomposition - do not actually solve

. . . $\frac{\text{some quadratic}}{x^3+x^2+3x+3}$

The denominator factors:
. . $x^3 + x^2 + 3x + 3 \;=\;x^2(x+1) + 3(x+1) \;=\;(x+1)(x^2+3)$

So we have: . $\frac{\text{some quadratic}}{(x+1)(x^2+3)} \;= \;\frac{A}{x+1} + \frac{Bx + C}{x^2 + 3}$

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