Question

find this.
∫x(x²+4)⁵ dx​

Answer 1

Answer:

Given

∫ (x−5) 45

(x+1) 43

1 dx

Substitute u= 4

x−2

dx=4(x−2) 43

du=4∫ u 2(u 4+3) 43

1du

Substitute v= u4u 4+3

, du= (u 4+3) 43

u 2 − u 24u 4+3

1dv=∫− 31dv=− 31∫dv=− 3v

=− 3u4u 4+3

=− 3u4 4u 4+3

=− 3 4x−244x+1

=− 34 ( x−2x+1 ) 41