Another substitiution question

[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

 

[arildno]

No, it is not right.
Split your integral as follows:
\int\frac{1+x}{1+x^{2}}dx=\int\frac{dx}{1+x^{2}}+\int\frac{xdx}{1+x^{2}}
maybe that helps.

 

[neutrino]

Split the first integral in two.

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

 

[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.

 

[radou]

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.

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.

 

[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

 

[arildno]

Why should it be problematic to do so?