# Thread: Indefinite integration with u-substitution

1. ## Indefinite integration with u-substitution

the first is (sqrt) ((x-1)/(x^5))

The first thing i did was distribute the x^5 and got:

(sqrt)(x^(-4) - x^(-5))...

I gave the value (x^(-4) - x^(-5)) for u and got:

du = -4x^(-3) + 5x^(-4). I am not sure if I am doing this right so far.

THe answer is (2/3)(1-(1/x))^(3/2) + C

The second is dz/(1 + e^z)

I used (1 + e^z) to equal u and du = e^z.

This is where I got confused.

the answer was: z - ln(1+e^z) + C... the z part baffled me more.

2. The Second Integral :

$
\int \frac{dz}{1+(e^z)}= \int \frac{(e^z)dz}{(1+(e^z))e^z} \rightarrow u=e^z \rightarrow du = dz e^z \rightarrow \int \frac{du}{(u+1)(u)}
$

$
\rightarrow \int \frac{du}{u} - \int\frac{du}{u+1} = ln u - ln (u+1) =ln e^z - ln ((e^z)+1) = z - ln((e^z)+1)
$

3. The Second Integral :

$
\int \frac{dz}{1+(e^z)}= \int \frac{(e^z)dz}{(1+(e^z))e^z} \rightarrow u=e^z \rightarrow du = dz e^z \rightarrow \int \frac{du}{(u+1)(u)}
$

Now we are going to seprate

$
\rightarrow \int \frac{du}{u} - \int\frac{du}{u+1} = ln u - ln (u+1) =z - ln ((e^z)+1)
$