# Thread: Indefinite Integral

1. ## Indefinite Integral

I'm trying to solve a problem that is for my final exam review and i'm quite stuck with these two problems..
1. \displaystyle \displaystyle \begin{align*} \int{x^3\sqrt{x^2 + 38}\,dx} \end{align*}
for this problem, i know the basics where i have to solve for the square first then solve for the indefinite.
I've tried to calculate it and this is my answer
could you please check it if it is right or wrong. Or if it's wrong could you explain it step by step for me.
Your help will be appreciated!

2.

this is the second problem and i don't have any clue on how to solve the problems.
Step by step help will be appreciated..
Thanks for everyone who will help cause I need this materials for next week's final exam!

2. ## Re: Indefinite Integral

Originally Posted by marmutgoreng
I'm trying to solve a problem that is for my final exam review and i'm quite stuck with these two problems..
1. x3(x2 +38) dx

for this problem, i know the basics where i have to solve for the square first then solve for the indefinite.
I've tried to calculate it and this is my answer
could you please check it if it is right or wrong. Or if it's wrong could you explain it step by step for me.
Your help will be appreciated!

2.

this is the second problem and i don't have any clue on how to solve the problems.
Step by step help will be appreciated..
Thanks for everyone who will help cause I need this materials for next week's final exam!
Please learn some LaTeX, it would make everything so much easier to read... Is the first one \displaystyle \displaystyle \begin{align*} \int{x^3\sqrt{x^2 + 38}\,dx} \end{align*}?

3. ## Re: Indefinite Integral

Originally Posted by Prove It
Please learn some LaTeX, it would make everything so much easier to read... Is the first one \displaystyle \displaystyle \begin{align*} \int{x^3\sqrt{x^2 + 38}\,dx} \end{align*}?
yes that's the exact question, I'm sorry,I'll edit the question right away.

4. ## Re: Indefinite Integral

First note that \displaystyle \displaystyle \begin{align*} \int{x^3\sqrt{x^2 + 38}\,dx} = \frac{1}{2}\int{x^2\sqrt{x^2 + 38}\,2x\,dx} \end{align*}

Let \displaystyle \displaystyle \begin{align*} u = x^2 + 38 \implies du = 2x\,dx \end{align*} and the integral becomes

\displaystyle \displaystyle \begin{align*} \frac{1}{2}\int{x^2\sqrt{x^2 + 38}\,2x\,dx} &= \frac{1}{2}\int{\left(u - 38\right)\sqrt{u}\,du} \\ &= \frac{1}{2}\int{u^{\frac{3}{2}} - 38u^{\frac{1}{2}}\,du} \\ &= \frac{1}{2}\left(\frac{u^{\frac{5}{2}}}{\frac{5}{2 }} - \frac{38u^{\frac{3}{2}}}{\frac{3}{2}}\right) + C \\ &= \frac{1}{2}\left[\frac{2}{5}\left(x^2 + 38\right)^{\frac{5}{2}} - \frac{76}{3}\left(x^2 + 38\right)^{\frac{3}{2}}\right] + C \\ &= \frac{1}{5}\left(x^2 + 38\right)^{\frac{5}{2}} - \frac{38}{3}\left(x^2 + 38\right)^{\frac{3}{2}} + C \end{align*}

So you were almost there, you just forgot to halve everything.

5. ## Re: Indefinite Integral

Originally Posted by Prove It
First note that \displaystyle \displaystyle \begin{align*} \int{x^3\sqrt{x^2 + 38}\,dx} = \frac{1}{2}\int{x^2\sqrt{x^2 + 38}\,2x\,dx} \end{align*}

Let \displaystyle \displaystyle \begin{align*} u = x^2 + 38 \implies du = 2x\,dx \end{align*} and the integral becomes

\displaystyle \displaystyle \begin{align*} \frac{1}{2}\int{x^2\sqrt{x^2 + 38}\,2x\,dx} &= \frac{1}{2}\int{\left(u - 38\right)\sqrt{u}\,du} \\ &= \frac{1}{2}\int{u^{\frac{3}{2}} - 38u^{\frac{1}{2}}\,du} \\ &= \frac{1}{2}\left(\frac{u^{\frac{5}{2}}}{\frac{5}{2 }} - \frac{38u^{\frac{3}{2}}}{\frac{3}{2}}\right) + C \\ &= \frac{1}{2}\left[\frac{2}{5}\left(x^2 + 38\right)^{\frac{5}{2}} - \frac{76}{3}\left(x^2 + 38\right)^{\frac{3}{2}}\right] + C \\ &= \frac{1}{5}\left(x^2 + 38\right)^{\frac{5}{2}} - \frac{38}{3}\left(x^2 + 38\right)^{\frac{3}{2}} + C \end{align*}

So you were almost there, you just forgot to halve everything.
thanks..
I've rechecked my equation and I realized that I forgot to halve everything.
Now I'm working on the 2nd problem and didn't even have a clue on how to do it..
Anyone can help?