1. ## Integrals

Can anyone help me out with these integrals step by step.

The integral from 0 to 1 of (sqrt(t)) / (t + 1)

Integral of (3x+2)/ x((x+2)^2)+16x

any help would be appreciated thanks

2. Originally Posted by poopforbrains
The integral from 0 to 1 of (sqrt(t)) / (t + 1)
$\displaystyle \int_0^1 \frac{\sqrt{t}}{t + 1}~dt$

Let $\displaystyle t = x^2 \implies dt = 2x~dx$:
$\displaystyle \int_0^1 \frac{\sqrt{t}}{t + 1}~dt = \int_0^1 \frac{\sqrt{x^2}}{x^2 + 1} \cdot 2x~dx$

$\displaystyle = 2 \int_0^1 \frac{x^2}{x^2 + 1}~dx$

Can you work with this one now?

-Dan

3. Originally Posted by poopforbrains
Integral of (3x+2)/ x((x+2)^2)+16x
I would like you to verify. This integral is correct?
$\displaystyle \int \frac{3x + 2}{x(x + 2)^2 + 16x}~dx$

-Dan

4. yup

5. $\displaystyle \int \frac{3x + 2}{x(x + 2)^2 + 16x}~dx$

$\displaystyle = \int \frac{3x + 2}{x^3 +4x^2 + 20x}~dx$

$\displaystyle = \int \frac{3x + 2}{x(x^2 + 4x + 20)}~dx$

Now do partial fractions:
$\displaystyle \frac{3x + 2}{x(x^2 + 4x + 20)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4x + 20}$

(The kid's crying. Gotta go!)

-Dan

6. mmm I got the 2nd one but the first is still giving me trouble....I dont know if I should substitue again

7. You don't need to set any substitution.

Note that

$\displaystyle \frac{{x^2 }}{{x^2 + 1}} = \frac{{x^2 + 1 - 1}}{{x^2 + 1}} = 1 - \frac{1}{{x^2 + 1}}.$

When integrating, the first term it's just $\displaystyle x,$ and the second one is the standard arctangent form.