Verified answer

letu=x^4+5 let du=4xdx

you forgot the (1/4)

int sqrt(x^4+5) * (x^3)dx

int (1/4) sqrt(u) du

based on sqrt(u) = u^(1/2)

(1/6)*u^(3/2)

based on (1/6)*(2/3)*u^(3/2)

(1/6)*(x^4+5)^(3/2)

Let u = 5 + x^4

du = 4x^3 dx

( 1 / 4 )du = x^3 dx

int [ (1 / 4)*u^(1 / 2 ) du ]

= ( 1 / 4 ) * int [ u^( 1 / 2) du ]

= ( 1 / 4 )( 2 / 3 )u^(3 / 2 ) + C

= ( 1 / 6)u^(3 / 2 ) + C

= ( 1 / 6 )(5 + x^4 )^(3 / 2 ) + C

well, it could be rewritten:

= ( 1 / 6 )(5 + x^4 ) sqrt ( 5 + x^4 ) + C

Dont alternative, divide the num via the denom and get two integrals: 4x/(x^2+1) +three/(x^2+1) in the first quintessential, discover that you've got a energy of x in the num and x^2 in the denom, so if you set u=x^2+1 here, you might have du=2x dx or x dx = du/2, and the first imperative becomes: 2 du/u = 2 log u = 2 log (x^2+1) the 2d critical is solely the indispensable of arctan x, so your outcome +3 arctanx