Question

integrate x^3 (x^2 +1)^(1/2) dx

Answer 1

\[\int\limits_{}^{} X^{3}(X^2+1)^{1/2}dx\]

Answer 2

Hmm you can solve this one either by using `Integration by Parts` several times, or by applying a `Trigonometric Substitution`. Are you comfortable with either of these methods?

Answer 3

I have tried both. x=atantheta seems to work easier than integration by parts
\[\int\limits\limits_{}^{} \tan^3\theta \sqrt{\tan^{2}\theta+1}\sec^2\theta dtheta\]
using the trig identity sec^2 -1= tan^2 and I get
\[\int\limits_{}^{} \sec^4 \theta \sec \theta \tan \theta d \theta - \sec^2 \theta \sec \theta \tan \theta \ \ d \theta \]
and then I solve and substitute \[\sec = \sqrt{1+x}/1 \]
and get \[(1/5) (1+x)^{5/2}-(1/3) (1+x)^{3/2} + C\]

It's just that the book and Wolfram Alpha have it very different:(

Answer 4

Hmm I think your answer is equivalent to that of Wolfram. Lemme see if I can jimmy some stuff around and make them match up. They look very close.

Answer 5

Yup it worked!

Answer 6

Oh I do see one little mistake, the inside of your brackets should be 1+x^2. Hmm

Answer 7

oh thanks. I haven't had calculus in 14 years and algebra in 15. That's my major problem with those little mistakes. Thanks for your help and sorry for the long reply!

Answer 8

\[\large \frac{1}{5}\left(x^2+1\right)^{5/2}-\frac{1}{3}\left(x^2+1\right)^{3/2}\]Factor \(\large \left(x^2+1\right)^{3/2}\) out of each term,\[\large \left(x^2+1\right)^{3/2}\left[\frac{1}{5}\left(x^2+1\right)-\frac{1}{3}\right]\]Factor \(\large \dfrac{1}{15}\) out of each term,\[\large \frac{1}{15}\left(x^2+1\right)^{3/2}\left[3\left(x^2+1\right)-5\right]\]

Answer 9

Then just a couple more steps to make them match. You seem to be on the right track! I know it can be frustrating sometimes since the book way way way oversimplifies things to save space.

Answer 10

Now I see it. THANKS!