Evaluating indefinite Integral

**kspkido** · November 26th 2012, 05:38 PM

Could someone please work me out in these problem, I did a substitution on Ln(x) but ended up with (u^2+1)/(u+1)...
here is the question find the indefinite integral of ((ln x)^2+1)/(xln(x)+3)....

**Prove It** · November 26th 2012, 05:48 PM

Are you sure it's not $\displaystyle \begin{align*} \int{\frac{\left( \ln{x} \right)^2 + 1}{x \left( \ln{x} + 3 \right)}\,dx} \end{align*}$ instead of $\displaystyle \begin{align*} \int{\frac{\left( \ln{x} \right)^2 + 1}{x\ln{x} + 3}\,dx} \end{align*}$ ?

**kspkido** · November 26th 2012, 06:40 PM

Oh I'm very sorry, it is actually (x ln x +3x)..... the 3 has 'x' with it..

**Prove It** · November 26th 2012, 06:55 PM

So it is what I thought.

I would write it as $\displaystyle \begin{align*} \int{\frac{\left( \ln{x} \right)^2 + 1}{\ln{x} + 3}\cdot \frac{1}{x}\,dx} \end{align*}$ .

Now this is just ONE possible substitution. I'm choosing it because I know that it will make the denominator the easiest to work with.

Let $\displaystyle \begin{align*} u = \ln{x} + 3 \implies du = \frac{1}{x} \, dx \end{align*}$ . Then your integral becomes

$\displaystyle \begin{align*} \int{\frac{\left( \ln{x} \right)^2 + 1}{\ln{x} + 3}\cdot \frac{1}{x} \,dx} &= \int{\frac{\left( u - 3 \right)^2 + 1}{u}\,du} \\ &= \int{\frac{u^2 - 6u + 9 + 1}{u}\,du} \\ &= \int{\frac{u^2 - 6u + 10}{u}\,du} \\ &= \int{u - 6 + \frac{10}{u}\,du} \\ &= \frac{u^2}{2} - 6u + 10\ln{|u|} + C \\ &= \frac{\left( \ln{x} + 3 \right)^2}{2} - 6\left( \ln{x} + 3 \right) + 10\ln{\left| \ln{x} + 3 \right|} + C \end{align*}$

**kspkido** · November 26th 2012, 09:00 PM

daym... I got the u^2-6u+10/u... but at that point I didn't know what to do.. so I tried to substitute other variables.. stupid me... Thank you very much...

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