u-subsitution?

Question

boundries 0 to 1
(x)/(x^2+1)

Answer 1

So you want to integrate:

Integral ( [x / (x^2 + 1)] dx ) {from 0 to 1}

I'm going to rearrange the terms in the integral so I can show you something.

Integral ( [1 / (x^2 + 1)] (x) dx ) {from 0 to 1}

Now, we're going to use substitutition with u = x^2 + 1, AND at the same time we're going to declare our new boundaries for integration. Note that it's currently 0 to 1, and your step to obtain the new bounds goes as follows:

Let u = x^2 + 1.

When x = 0, u = 0^2 + 1 = 1. When x = 1,
u = 1^1 + 1 = 2. Thus, our new bounds are from 1 to 2.

Then du = 2x dx. Multiplying both sides by (1/2), we get
(1/2) du = x dx

See how the tail end of our integral is x dx? That's precisely what we're going to replace in our substitution, with (1/2) du.

Our integral then becomes:

Integral ( [1 / u] (1/2) du ) {from 1 to 2}

It's always a good idea to pull out constants; in our case, we have 1/2 inside the integral. We want to pull this OUT of the integral. As a side note, the handy thing about derivatives, integrals, and limits, is that you can ALWAYS solve for the integral while ignoring the constant.

(1/2) * Integral ( (1/u) du ) {from 1 to 2}

Now, integrate normally.

(1/2) [ ln|u| ] {from 1 to 2}

Now, insert the integration bounds.

(1/2) [ ln(2) - ln(1) ]

Note that ln(1) = 0, so we have

(1/2) ln(2)

Answer 2

u = (x^2 + 1)
du = 2x dx

So you have to balance the 2 with a 1/2 in front of the integral in the u-substitution...

Therefore the integral becomes:

1/2 integral[0,1] du / u

As you learned, the integral of du/u = ln |u| + c

So now just re-substitute:

1/2 ln |x^2 + 1| evaluatated from 0 to 1
= 1/2 ln|0 + 1| - 1/2 ln|1 + 1|
= 0 - (1/2) ln 2
= -(1/2) ln 2