# Thread: Two (Hopefully the last two!) Integration Questions

1. ## Two (Hopefully the last two!) Integration Questions

These questions share a similar theme.

1) Integral of (x-1)dx/(x^2 - 4x + 5)

I have tried dividing this problem into two fractions, without result. I thought maybe difference of squares somehow on the bottom, x^2 - 4x + 4 + 1, but what then?

2) xdx/x^4 + x^2 + 1

I set u = x^2
du = 2x
= 1/2 Integral of du/u^2+u+1

And I STILL can't solve it! It is in a really simple form now, 1/u^2 + u + 1, what am I doing wrong?

Seems like it would be obvious, but it's evading me.

Sigh. Why is this not obvious to me? V. frustrating

2. $\int\frac{x-1}{x^2-4x+5}dx$

Well you know that:

$x^2-4x+5=x^2-4x+4 +1 = (x-2)^2+1$

so:

$\int\frac{x-1}{(x-2)^2+1}$

so let's let u = x-2

so du= dx and u+1=x-1

$\int\frac{u+1}{u^2+1}$

Then I would do a second substitution with trigonometry:

let $u = tan\theta$

$du = \frac{1}{cos^2\theta}d\theta$

$\int\frac{tan\theta+1}{tan^2\theta +1}*\frac{1}{cos^2\theta}d\theta$

$\int\frac{tan\theta+1}{\frac{1}{cos^2\theta}}*\fra c{1}{cos^2\theta}d\theta$

$\int (tan\theta+1)d\theta$

3. Ah, it was the difference of squares trick. I don't feel confident using it, it always seems suspicious to me!

4. Well the truth is that I went through calculus I, II and III with only power rule, product rule, derivative of sine and cosine memorized.

Something useful to remember with trig substitutions though:

$sin^2\theta+cos^2\theta=1$

divide by cos^2 on both sides:
$\frac{sin^2\theta}{cos^2\theta}+\frac{cos^2\theta} {cos^2\theta}=\frac{1}{cos^2\theta}$

$tan^2\theta+1=sec^2\theta$

$sin^2\theta+cos^2\theta=1$

divide by sin^2 on both sides:
$\frac{sin^2\theta}{sin^2\theta}+\frac{cos^2\theta} {sin^2\theta}=\frac{1}{sin^2\theta}$

$1+cot^2\theta=csc^2\theta$

And you don't have to memorize is it tan^2 -1 = csc^2 ? or whatever just straight and simple.

5. $\frac{u+1}{u^{2}+1}=\frac{u}{u^{2}+1}+\frac{1}{u^{ 2}+1}.$