# Integral help

this should be easy, but i'm struggling. i'm trying to help a friend solve:

integral(5x^3(1+x^2)^1/2), sorry i don't know how to do the integral sign.

i did a u sub with u=1+x^2, du=2xdx
and u-1=x^2

so:

(5/2)integral((u-1)(u)^1/2)

i get (1+x^2)^5/2-10/6(1+x^2)^3/2 +C

the answer says it should be 15/8(1+x^2)^4/3+C

i'm not sure what i have done wrong.

********THIS WAS WRONG*********

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Umm. hmph. Thats weird.

wow. that is unlike either answer i posted. i guess i should have stated that the only special way they have learned to solve integrals is with a u sub. did you see what exactly i did wrong in the way i tried to solve it? sometimes one needs an extra pair of eyes to find the error.

Wait a minute.

$$t = tan(u)$$

$$dt = tan(u)sec(u)du$$

is incorrect. dt = sec ^2 (u) du

Oh I see where I messed up.

$$\int{5x^3\sqrt{1+x^2}}{dx}$$

You can do this with the trig substitution $$x = tan(u)$$
Then $$dx = sec^2(u)du$$

The integral becomes:

$$5\int{tan^3(u)\sqrt{1+tan^2(u)}sec^2(u)}{du}$$

$$1 + tan^2(u) = sec^2(u)$$

$$5\int{tan^3(u)\sqrt{sec^2(u)}sec^2(u)}{du}$$

$$5\int{tan^3(u)sec^3(u){du}$$

$$5\int{tan^2(u)sec^2(u)sec(u)tan(u)}{du}$$

$$5\int{(1-sec^2(u))sec^2(u)sec(u)tan(u)}{du}$$

Substituting $$t = sec(u), dt=sec(u)tan(u)du$$

$$5\int{(1-t^2)t^2dt = 5\int{t^2-t^4}{dt}$$

$$\frac{5t^3}{3}-\frac{5t^5}{5}$$

I butchered that first one but I don't see anything wrong with this one.

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$$\frac{5}{2}\int(u-1)\sqrt{u}{du}$$

sorry, but i'm not sure how your answer is equivalent to the answer she was given ie)15/8(1+x^2)^4/3 +C. also, i guess i wasn't clear in my last post. they haven't done trig subs, by parts, partial fractions, etc, nothing like that yet. all they have done is u subs. this isn't even my question, but it is irritating me beyond belief. what is wrong with what i did?

$$\frac{5}{2}\int(u-1)\sqrt{u}{du} = \frac{5}{2}\int{u^{3/2}-u^{1/2}}{du}$$

$$\frac{5}{2}( \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}) = u^{5/2}-\frac{5u^{3/2}}{3}$$

Whozum,

did not mean to find little mistakes. But I did this one and got the same results as Jidon. You have made couple of mistakes in the final two steps.

tan^u = sec^2 u -1 = t^2 - 1

$$5\int{(t^2 -1)t^2dt = 5\int{t^4-t^2}{dt}$$

$$t^5-\frac{5t^3}{3}$$

Jaidon: Your way is much simpler than trig sub. I don't see any thing wrong. Your book answer seems to be way off from what we are getting. Are you sure about that the given answer?

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I dont see anything wrong with what you did, but the thing is when I derive their answer in maple I dont get anything near the original integral.

Its ok gamma, better you find them than we let them run.

does anyone agree with me that maybe the answer given by my friend's sheet is incorrect?

You said the answer was 15/8(1+x^2)^4/3

$$\frac{15}{8} (1+x^2)^{4/3}$$

Deriving that gives $$\frac{15}{8}*\frac{4}{3}(1+x^2)^{1/3}*2x$$ by the chain rule.

$$5x(1+x^2)^{1/3}$$

just wanted to let anyone who helped out with this know that my friend and i figured out, by looking at the derivative of the answer the prof gave, that there is a typo in the original question. it should not be 5x^3 (1+x^2)^1/2, but 5x (1+x^2)^1/3. it was written in such a way that i don't know anyone who would have read it as it should be. thought that was quite funny and just wanted to let my helpers know. thanks again for the help

jaidon

In that case I integrated it right :)