Question

need help with evaluating this integral.... (x^9)/(sqrt(3+x^5)) dx.....

Answer 1

I would recommend a u substitution of u = x^5+3

Answer 2

ok i started that....

Answer 3

How far did you get?

Answer 4

then i got (1/5)du = x^4dx ....

Answer 5

Ok, though I prefer dx = du/(5x^4)

Answer 6

ok.... then

Answer 7

So then, what is x^5 in terms of u?

Answer 8

i dont know what you mean by that

Answer 9

u = x^5 + 3 so x^5 = ?

Answer 10

x^5 = u - 3 right?

Answer 11

u - 3

Answer 12

ok. Good. Now start substituting in your integral

Answer 13

so whats the x^9?... the denominator will be sqrt3?

Answer 14

The denominator will be \(\sqrt{u}\)

Answer 15

thats what i meant.///

Answer 16

the dx will be \[\frac{1}{5x^4}du\]

Answer 17

So what's left in the numerator( after you simplify)

Answer 18

idk.......

Answer 19

Cmon.

\[\int \frac{x^9}{\sqrt{x^5+3}}dx = \int \frac{x^5 * x^4}{\sqrt{x^5+3}}dx \]
\[\text{Let }u = x^5+3 \]\[\implies x^5 = u-3 \]
\[\implies 5x^4dx = du \implies dx = \frac{1}{5x^4}du\]
Therefore
\[\int \frac{x^5 * x^4}{\sqrt{x^5+3}}dx = \int \frac{(u-3)*x^4}{\sqrt{u}} (\frac{1}{5x^4})du\]

Answer 20

And we can cancel the x^4 from top and bottom. Then split the fraction into two different fractions
\[\frac{u}{\sqrt{u}} - \frac{3}{\sqrt{u}}\]

Answer 21

and integrate each one separately.

Answer 22

what happened to the 5 in the faraction

Answer 23

yes there should be a factor of 5 outside the integral , he most likely didnt mention it because any integral of a constant multiple is just the integral times that constant