# Another substitiution question

1. Oct 22, 2006

### sapiental

Evaluate the integral

$$\int \frac {1+x}{1+x^2}dx$$

i let u = 1+x
du = dx

so
$$\int \frac {dx}{u^2}du$$

or

$$\int u^-2du$$

$$\frac {-1}{u} + C$$

im not sure if this is right, the only other thing I can think of is to let u = (1+x)/(1+x^2) and then use the quotient rule.

Any feedback is much appreciated.

Thanks

2. Oct 22, 2006

### arildno

No, it is not right.
$$\int\frac{1+x}{1+x^{2}}dx=\int\frac{dx}{1+x^{2}}+\int\frac{xdx}{1+x^{2}}$$
maybe that helps.

3. Oct 22, 2006

### neutrino

Split the first integral in two.

Btw, u^2 = (1+x)^2 != 1+x^2

4. Oct 22, 2006

### sapiental

hmm, that makes sense. Would I evaluate them like this then?

for the first integral

u = 1 + x
du = dx

and for the second

u = 1 + x^2
then du/2 = xdx

so then

$$\int \frac {1+x}{1+x^2}dx$$ = $$\int \frac {du}{u^2}$$ + $$\int \frac {du}{u}$$

$$\frac {-1}{u}$$ + ln|u| + C

thanks again.

Last edited: Oct 22, 2006
5. Oct 22, 2006

The second one is $$\frac{1}{2} \int \frac{du}{u}$$. As for the first one, note that $$u^2 = (1+x)^2 \neq 1+ x^2$$.

6. Oct 22, 2006

### sapiental

ohhh, I oversee too many details.

Can I just leave the integral $$\int \frac {dx}{1+x^2}$$

the way it is and just take the antiderivative to be tan^-1(x) + C

7. Oct 22, 2006

### arildno

Why should it be problematic to do so?