# Use a substitution to compute the integral

1. Mar 11, 2014

### Neek 007

1. The problem statement, all variables and given/known data
Going over past exam problems, stuck on this one. Attached

Calc 2, topics include for this exam integration techniques, such as partial fractions, improper integrals, trig sub, and series.

Question reads: Use a substitution to compute: (see attached)

2. Relevant equations

3. The attempt at a solution

I tried partial fractions integration.

1. dx/x(1 + sqrt(x)

2. A/x + B/(1+sqrt(x) = 1/x(1+sqrt(x))

Solved for A, A= 1 B = -1

Resulted in:

1/x - 1/1+sqrt(x)

I integrated each part:

ln(abs(x)) - (ln(abs(1+sqrt(x)))*(1/2sqrt(x))

I am not sure if this was correct.

#### Attached Files:

• ###### daum_equation_1394583054024.png
File size:
750 bytes
Views:
56
2. Mar 11, 2014

### ehild

$\frac{1}{x(1+\sqrt{x}) }$ is not equal to 1/x - 1/(1+sqrt(x)).

Use the substitution √x = u first.

ehild

3. Mar 11, 2014

### Neek 007

u = sqrt(x)
du = 1/2sqrt(x)

x = u^2

∫1/(u^2 + u^3)

∫1/(u^2(1+u))

Then I use partial fractions decomposition

A= 1
B = -1

∫1/u^2 - ∫1/(1+u)

substitute in u

∫1/x - ∫1/(1+sqrt(x))

ln(abs(x)) - ln(abs(1+sqrt(x)))

4. Mar 11, 2014

### ehild

That is wrong...

Determine dx in terms of u and du. Without dx or du, it is not an integral!

ehild

5. Mar 11, 2014

### Simon Bridge

Did you forget to sustitute for dx?

Make it easier for you: Define $x=u^2 \implies dx=?$

6. Mar 11, 2014

### Neek 007

so how does this sound:

2u/(u^2(1+u)) du

Then I complete the partial fraction decomposition?

7. Mar 11, 2014

### Curious3141

Simplify by cancelling first, then do the partial fraction decomposition.