Question

How do you integrate this using substitution ?
∫x^3(5+x^2)^1/2dx

Answer 1

\[\int x^{3}\sqrt{5+x^2}dx\]

Answer 2

let u = \(x^2\) =>du/dx = \(2x\)

Answer 3

hmm

Answer 4

\[∫ 1/(x²+4)^3/2 dx = 1/4 \sin \quad u + C\]
\[∫ 1/(x²+4)^3/2 dx = x/(4√(x²+4)) + C \]

Answer 5

\[\int\limits x^{3} \sqrt{5+u} *\frac{du}{2x} => \int\limits x*x^{2} \sqrt{5+u} *\frac{du}{2x} => \int\limits x*u\sqrt{5+u} *\frac{du}{2x} \]
\[ =>\frac{1}{2} \int\limits u\sqrt{5+u} \]

Answer 6

Are you able to do it now?

Answer 7

@lgbasallote: What's wrong?

Answer 8

nothing wrong with that

Answer 9

@Mimi_x3 Why did you write u in place of x^2 ? :/

Answer 10

nvm got it .thanks alot :)

Answer 11

np

Answer 12

sorry one more question @Mimi_x3 , why did you set u = x^2 . Isnt it always the inside function so in this case u = 5+x^2 ?

Answer 13

I would do it in this way...
\[\int x^3 \sqrt{5+x^2} dx\]Let u = 5+x^2
du = 2x dx ; x^2 = u-5
\[\int x^3 \sqrt{5+x^2} dx\]\[=\frac{1}{2}\int x^2 \sqrt{5+x^2} (2xdx)\]\[=\frac{1}{2}\int (u-5) \sqrt{u} du\]\[=\frac{1}{2}\int u^{\frac{3}{2}}-5u^{\frac{3}{2}} du\]\[=...\]

Answer 14

*for the last step before ..., it is
\[=\frac{1}{2} \int u^{\frac{3}{2}}-5u^{\frac{1}{2}}du\]

Answer 15

thankyou :D

Answer 16

yeah but you have to change the variable..
any way would work