Question

help me integrate this

Answer 1

\[\int\limits_{1}^{\sqrt{7}}x/\sqrt[3]{x^2+1}dx\]

Answer 2

pls

Answer 3

Try u-sub
u=x^2+1
du=2xdx
du/2=xdx

Answer 4

\[\frac{1}{2}\int\limits \frac{du}{\sqrt[3]{u}}\]

Answer 5

dont i have to rewrite it first

Answer 6

?

Answer 7

\[\bf \int\limits_{1}^{\sqrt7}x(x^2+1)^{-1/3}dx\]We can do a u-substituion for x^2+1.\[\bf \int\limits_{1}^{\sqrt7}\cancel{x}(u)^{-1/3}\frac{ du }{ 2\cancel{x} }=\int\limits_{1}^{\sqrt7}(u)^{-1/3}du=\int\limits_{2}^{8}(u)^{-1/3}du= \frac{ 3 }{ 2 }(8^{2/3}-2^{2/3})=..?\]
@yomamabf

Answer 8

I skipped a couple steps for the antiderivative there.