Question

Evaluate the indefinite integral of x^(3) (x^2+49)^1/2 dx

Answer 1

Will most likely need to do a trig substitution. Try letting:\[x=\tan\theta\]

Answer 2

I did u-substitution and let u^2 = x^2+49

Answer 3

I'm not quite sure if that will lead to a solution.

Answer 4

well I don't know what I'm doing haha

Answer 5

integral by u-sub will works

u = x^2 + 49 --------> x^2 = u - 49
du = 2x dx
1/2 du = x dx

now see ur integral becomes
int x^3 (x^2 + 49)^1/2 dx
= int x x^2 (x^2 + 49)^1/2 dx
= 1/2 int (u-49) u^1/2 du
= 1/2 int (u^(3/2) - 49u^1/2) du

Answer 6

that should make easier for u, now

Answer 7

now do you plug x^2+49 back in for u?

Answer 8

after integration, yes :)

Answer 9

so now I'll have to take the anti derivative of u^3/2 and 49u^1/2?

Answer 10

yup

Answer 11

I got 3/5u^5/2 and 98/3u^3/2

Answer 12

that's 3/5 or 2/5 ?

Answer 13

whoops 2/5 i meant

Answer 14

ok, u are right
dont forget multiply by 1/2 and subtitute back u = ...
endly + c

Answer 15

would it be: 1/2[2/5(x^2+49)^3/2 - 49(x^2+49)1/2] +C?

Answer 16

1/2 [2/5(x^2+49)^5/2 - 98/3(x^2+49)^3/2] +C
= 1/5(x^2+49)^5/2 - 49/3(x^2+49)^3/2 +C

Answer 17

is that with the 1/2 multiplied through?

Answer 18

look my coments before above
= 1/2 int (u-49) u^1/2 du
= 1/2 int (u^(3/2) - 49u^1/2) du
1/2 * .........
yeah, after integration we have to multiply that by 1/2

Answer 19

Can it be simplified further after multiplying by 1/2?

Answer 20

yeah, it have simplied above
1/2 [2/5(x^2+49)^5/2 - 98/3(x^2+49)^3/2] +C
= 1/5(x^2+49)^5/2 - 49/3(x^2+49)^3/2 +C

Answer 21

Awesome!! thank you for helping me with the problem!