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Question

(5x+3)2x1dx

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Solution

Let I=(5x+3)2x1dx

Let 5x+3=λ(2x1)+μ
Comparing the coefficients, we get

2λ=5λ=52
λ+μ=352+μ=3
μ=3+52=6+52=112
=(λ(2x1)+μ)2x1dx
=(52(2x1)+112)2x1dx

I=(5x+3)2x1dx
=52(2x1)1+12dx+112(2x1)12dx
=52(2x1)32dx+112(2x1)12dx

=52(2x1)32+12(32+1)+112(2x1)12+12(12+1)+c
=52(2x1)522(52)+112(2x1)322(32)+c
=52(2x1)525+112(2x1)323+c
=12(2x1)52+116(2x1)32+c

=16(2x1)32[3(2x1)+11]+c
=16(2x1)32[6x3+11]+c
=16(2x1)32[6x+8]+c
=16(2x1)32[2(3x+4)]+c
=26(2x1)32(3x+4)+c
=13(2x1)32(3x+4)+c

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