Question

integral of sqrt(x)/(1+x^3)

Answer 1

u=x^3/2

Answer 2

i was gonna try using integration by parts

Answer 3

but that isnt working

Answer 4

did you use the u substitution?

Answer 5

how did you know to use u=x^(3/2)?

Answer 6

?

Answer 7

diff of x^(3/2) is 3/2 * sqrt(x)

Answer 8

\[\int\frac{\sqrt x}{1+x^3}~dx\]
Let \(u=x^{3/2}~\Rightarrow~du=\dfrac{3}{2}x^{1/2}~dx\). (i.e. \(\dfrac{2}{3}du=\sqrt x~dx\).)
\[\frac{2}{3}\int\frac{du}{1+\left(u^{2/3}\right)^3}\\
\frac{2}{3}\int\frac{du}{1+u^2}\]
Carry on with a trig sub.

Answer 9

thanks. Its hard to figure out where to start sometimes with integrals, but im starting to figure out patterns as to which technique to use for certain forms of them

Answer 10

you're welcome!

Answer 11

Now can you explain WHY u =x^(3/2)?

Answer 12

because x^2 raised to the 3/2 power gives me x^3.

Answer 13

Oh I see... you're trying to rework it into the definition of tangent inverse... \(\large \int \cfrac{1}{1+x^2}dx\)

Answer 14

yes, something like that. :)

Answer 15

but i am ultimately trying to use trig substitution to get it into that form.

Answer 16

Are you still stuck?

Answer 17

i figured it out actually. Thanks though :)

Answer 18

\begin{align}
\large \int\frac{\sqrt x}{1+x^3}~dx &\\
\\
& \text{let u}= x^{3/2} \iff x=u^{2/3} \ , \ du = \frac{3}{2} x^{1/2}dx \rightarrow \frac{2}{3}du = \sqrt{x} dx \\
\\
&= \large \cfrac{2}{3}\int \frac{1}{1+(u^{2/3})^3}du \\
&= \large \cfrac{2}{3}\int \frac{1}{1+u^{(2/3) \cdot 3}} dx \\
&= \large \cfrac{2}{3}\int \frac{1}{1+u^2}du \\
&= \large \cfrac{2}{3} \tan^{-1}(u) +c \\
&= \large \frac{2}{3} \tan^{-1} (x^{3/2}) +c
\end{align}