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∫7+4x+5x2(2x+3)32dx

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Question

`int (7 + 4x + 5x^2)/(2x + 3)^(3/2) dx`

Sum
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Solution

Let `I = int(5x^2 + 4x +7)/(2x + 3)^(3/2) dx`

Put 2x + 3 = t2    ...(i)

Differentiating w.r.t. x, we get

2dx = 2t dt

∴ dx = t dt

From (i), we get

`x = (t^2 - 3)/2`

∴ `I = int (5((t^2 - 3)/2)^2 + 4((t^2 - 3)/2) + 7)/(t^2)^(3/2) * t dt`

`I = int (5((t^4 - 6t^2 + 9)/4) + 2t^2 - 6 + 7)/t^3 * t dt`

`I = int (5t^4 - 30t^2 + 45 + 8t^2 + 4)/(4t^3) * t dt`

`I = int (5t^4 - 22t^2 + 49)/(4t^2) dt`

`I = 5/4 int t^2 dt - 22/4 int dt + 49/4 int t^(-2) dt`

`I = 5/4 * t^3/3 - 22/4 t + 49/4 * t^(-1)/(-1) + c`

`I = 5/12t^3 - 11/2t - 49/(4t) + c`

∴ `I = 5/12(2x + 3)^(3/2) - 11/2 sqrt(2x + 3) - 49/4 * 1/sqrt(2x + 3) + c`

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Chapter 2.3: Indefinite Integration - Short Answers II

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