Use a substitution to compute the integral

[Neek 007]

Homework Statement

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)

Homework Equations

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.

 

[ehild]

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

Use the substitution √x = u first.


ehild

 

[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)))

Answer: ln((x/(1+sqrt(x)))

 

[ehild]

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

That is wrong...

x = u^2

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

ehild

 

[Simon Bridge]

Did you forget to sustitute for dx?

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

 

[Neek 007]

so how does this sound:2u/(u^2(1+u)) du

Then I complete the partial fraction decomposition?

 

[Curious3141]

so how does this sound:


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

Then I complete the partial fraction decomposition?

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