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Question
Integrate the following functions w.r.t. x : `(x + 1) sqrt(2x^2 + 3)`
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Solution
Let I = `int (x + 1)sqrt(2x^2 + 3)`
Let x + 1 = `"A"[d/dx (2x^2 + 3)] + "B"`
= A (4x) + B
= 4Ax + B
Comparing the coefficients of and constant on both sides, we get
4A = 1, B = 1
∴ A = `(1)/(4), "B"` = 1
∴ x + 1 = `(1)/(4)(4x) + 1`
∴ I = `int [1/4 (4x) + 1]sqrt(2x^2 + 3).dx`
= `(1)/(4) int 4x sqrt(2x^2 + 3).dx + int sqrt(2x^2 + 3).dx`.
= I1 + I2
In I1 = put 2x2 + 3 = t
∴ 4x.dx = dt
∴ I1 = `(1)/(4) int t^(12).dt`
= `(1)/(4)(t^(3/2)/(3/2)) + c_1`
= `(1)/(6)(2x^2 + 3)^(3/2) + c_1`
I2 = `int sqrt(2x^2 + 3).dx`
= `sqrt(2) int sqrt(x^2 + 3/2).dx`
= `sqrt(2)[x/2sqrt(x^2 + 3/2) + ((3/2))/(2)log|x + sqrt(x^2 + 3/2)|] + c_2`
= `sqrt(2)[x/2sqrt(x^2 + 3/2) + (3)/(4)log|x + sqrt(x^2 + 3/2)|] + c_2`
∴ I = `(1)/(6)(2x^2 + 3)^(3/2) + sqrt(2)[x/2 sqrt(x^2 + 3/2) + (3)/(4) log|x + sqrt(x^2 + 3/2)|] + c`, where c = c1 + c2.
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