Question

integrate by parts: ⌡(x^3)/sqrt((x^2)+1))

Answer 1

\(\huge \int \frac{x^3}{\sqrt{x^2+1}}dx \)

let:
\(\large u=x^3 \) and \(\large dv=\frac{1}{\sqrt{x^2+1}}dx \)
so
\(\large du=3x^2dx \) and \(\large v=arctanx \)

so
\(\large \int \frac{x^3}{\sqrt{x^2+1}}dx=v\cdot v - \int v du \)
=\(\large (x^3)(arctanx) - \int arctanx \cdot 3x^2dx \)

looks like you'll have to carry it out again... and again...

Answer 2

ohhh ok. i see what i did wrong

Answer 3

usually choose the u to be the one you take derivative so that the degree will be less after you take the derivative.

Answer 4

yeah i got that part but i set dv\[=x(x ^{2}+1)^{\frac{ 1 }{ 2 }}\] and v=\[=(x ^{2}+1)^{\frac{ 1 }{ 2 }}\]

Answer 5

sorry..yes its to -1/2