# Thread: [SOLVED] Indefinite Integration by Sub

1. ## [SOLVED] Indefinite Integration by Sub

Integrate:

a) $\displaystyle \int \frac{2xln(x^2 + 1)}{x^2 + 1}\, dx$

$\displaystyle Let\, u = x^2 + 1, du = 2x\, dx$

$\displaystyle = \int \frac{2xln(x^2 + 1)}{u} \frac{1}{2x}\, du$

$\displaystyle = \int ln(x^2 + 1) u^{-1}\, du$

$\displaystyle = ln(x^2 + 1) (ln(x^2 + 1)) + c$

$\displaystyle = [ln(x^2 + 1)]^2 + c$

answer is $\displaystyle \frac{1}{2} [ln(x^2 + 1)]^2 + c$, where does 1/2 come from??

b) $\displaystyle \int \frac {x}{2x + 1}\, dx = x(2x + 1)^{-1}\, dx$

$\displaystyle Let \, u = 2x + 1\, du = 2\, dx$

$\displaystyle = \frac{1}{2} \int x(u^{-1})\, du$

$\displaystyle = (\frac{1}{2})(\frac{1}{2})(x^2)(ln(2x + 1)) + c$

$\displaystyle = \frac{1}{4}x^2 (ln|2x + 1|) + c$

answer is $\displaystyle \frac{1}{2}x - \frac{1}{4}(ln|2x + 1|) + c$, what did I do wrong?

2. Originally Posted by Macleef
Integrate:

a) $\displaystyle \int \frac{2xln(x^2 + 1)}{x^2 + 1}\, dx$

$\displaystyle Let\, u = x^2 + 1, du = 2x\, dx$

$\displaystyle = \int \frac{2xln(x^2 + 1)}{u} \frac{1}{2x}\, du$

$\displaystyle = \int ln(x^2 + 1) u^{-1}\, du$

$\displaystyle = ln(x^2 + 1) (ln(x^2 + 1)) + c$

$\displaystyle = [ln(x^2 + 1)]^2 + c$

answer is $\displaystyle \frac{1}{2} [ln(x^2 + 1)]^2 + c$, where does 1/2 come from??

b) $\displaystyle \int \frac {x}{2x + 1}\, dx = x(2x + 1)^{-1}\, dx$

$\displaystyle Let \, u = 2x + 1\, du = 2\, dx$

$\displaystyle = \frac{1}{2} \int x(u^{-1})\, du$

$\displaystyle = (\frac{1}{2})(\frac{1}{2})(x^2)(ln(2x + 1)) + c$

$\displaystyle = \frac{1}{4}x^2 (ln|2x + 1|) + c$

answer is $\displaystyle \frac{1}{2}x - \frac{1}{4}(ln|2x + 1|) + c$, what did I do wrong?
When you're putting substitutions on integral you're not subst. everything, you're working with both x and u. When you substitute something with u, you have to express everthing inside of integral with u.
In that case in a) after sub. the integral would be. $\displaystyle \int{\frac{\ln{u} du}{u}}.$Here substitute again $\displaystyle \ln{u}=t \Rightarrow \frac{du}{u}=dt$
so the integral will be $\displaystyle \int{t dt}= \frac{t^2}{2}=\frac{ln^2{u}}{2}= \frac{(\ln{x+1})^2}{2}+c$
The same mistake in part b).

3. Originally Posted by javax
When you're putting substitutions on integral you're not subst. everything, you're working with both x and u. When you substitute something with u, you have to express everthing inside of integral with u.
In that case in a) after sub. the integral would be. $\displaystyle \int{\frac{\ln{u} du}{u}}.$Here substitute again $\displaystyle \ln{u}=t \Rightarrow \frac{du}{u}=dt$
so the integral will be $\displaystyle \int{t dt}= \frac{t^2}{2}=\frac{ln^2{u}}{2}= \frac{(\ln{x+1})^2}{2}+c$
The same mistake in part b).
I don't understand where "t" comes from... why is lnu = t?

also i followed your advice to turn every x variable to u for the second question and still not getting it...

$\displaystyle \int \frac{u}{2u + 1} du$

$\displaystyle \int 2u^2 + u du$

$\displaystyle \frac{2}{3}(x)^3 + \frac{1}{2}(x )^2 + c$

4. Because we want to solve it by substitution...That is the correct substitution because when you differentiate $\displaystyle \ln{u}=t$ it becomes

$\displaystyle \frac{du}{u}=dt$ so now the expression$\displaystyle \frac{du}{u}$ under integral you'll sub. with $\displaystyle dt$. When you finish the integral you have to get back the substitutions.

In part b). after doing those substitutions you'll have

$\displaystyle 2x+1 = u$

$\displaystyle x=\frac{u-1}{2}$

$\displaystyle dx=\frac{du}{2}$

now the integral will be

$\displaystyle \int{\frac{\frac{u-1}{2}}{u} \frac{1}{2}du}$
$\displaystyle =\frac{1}{4}\int{\frac{u-1}{u}du} = \frac{1}{4}\int{du}-\frac{1}{4}\int{\frac{du}{u}} = ...$now I believe you can finish it.

5. a) $\displaystyle \int \frac{2xln(x^2 + 1)}{x^2 + 1}\, dx$

note ...

$\displaystyle \int \frac{2xln(x^2 + 1)}{x^2 + 1}\, dx = \int \ln(x^2+1) \cdot \frac{2x}{x^2+1} \, dx$

$\displaystyle u = \ln(x^2 + 1)$

$\displaystyle du = \frac{2x}{x^2+1} \, dx$

$\displaystyle \int u \, du = \frac{u^2}{2} + C = \frac{1}{2}[\ln(x^2+1)]^2 + C$

b) $\displaystyle \int \frac {x}{2x + 1}\, dx = x(2x + 1)^{-1}\, dx$

$\displaystyle u = 2x+1$

$\displaystyle x = \frac{u-1}{2}$

$\displaystyle du = 2\, dx$

$\displaystyle \frac{1}{2}\int \frac{u-1}{2u} \, du$

$\displaystyle \frac{1}{4} \int 1 - \frac{1}{u} \, du$

$\displaystyle \frac{1}{4}(u - \ln|u|) + C$

$\displaystyle \frac{1}{4}[(2x+1) - \ln|2x+1|] + C$

$\displaystyle \frac{x}{2} + \frac{1}{4} - \frac{1}{4}\ln|2x+1| + C$

$\displaystyle \frac{x}{2} - \frac{1}{4}\ln|2x+1| + C$