If I=∫dx2ax+x23/2, then I is equal to

Write 2ax+x2=(x+a)2−a2, and put x+a=asec⁡θ, so that dx=asec⁡θtan⁡θdθ∴I=∫asec⁡θtan⁡θa3tan3⁡θdθ=1a2∫cos⁡θsin2θdθ=−1a2sin⁡θ+C=−1a2sec⁡θtan⁡θ+C=−1a2x+a2ax+x2+C