Let I=∫(5x+3)√2x−1dx
Let 5x+3=λ(2x−1)+μ
Comparing the coefficients, we get
⇒2λ=5⇒λ=52
⇒−λ+μ=3⇒−52+μ=3
⇒μ=3+52=6+52=112
=∫(λ(2x−1)+μ)√2x−1dx
=∫(52(2x−1)+112)√2x−1dx
∴I=∫(5x+3)√2x−1dx
=52∫(2x−1)1+12dx+112∫(2x−1)12dx
=52∫(2x−1)32dx+112∫(2x−1)12dx
=52(2x−1)32+12(32+1)+112(2x−1)12+12(12+1)+c
=52(2x−1)522(52)+112(2x−1)322(32)+c
=52(2x−1)525+112(2x−1)323+c
=12(2x−1)52+116(2x−1)32+c
=16(2x−1)32[3(2x−1)+11]+c
=16(2x−1)32[6x−3+11]+c
=16(2x−1)32[6x+8]+c
=16(2x−1)32[2(3x+4)]+c
=26(2x−1)32(3x+4)+c
=13(2x−1)32(3x+4)+c