∫y3√2y2+1dy
=∫y2⋅√2y2+1⋅ydy
Let
u=√2y2+1
u2=2y2+1
y2=12(u2−1)
2ydy=12(2udu)
ydy=12udu
Hence,
∫y3√2y2+1dy
=∫12(u2−1)⋅u⋅12udu
=14∫(u4−u2)du
=14(u55−u33)+C
=14(3u5−5u315)+C
=160u3(3u2−5)+C
Revert to original variable of integration
∫y3√2y2+1dy
=160(√2y2+1)3[3(√2y2+1)2−5]+C
=160(2y2+1)3/2[3(2y2+1)−5]+C
=160(2y2+1)3/2[6y2+3−5]+C
=160(2y2+1)3/2(6y2−2)+C
=130(2y2+1)3/2(3y2−1)+C ← answer