∫ {x-1}{(x+1){x³+x²+x}} dx =

Correct Answer is : 2 tan⁡−1(1+x+x2x)^{-1}≤ft({{1+x+x²}{x}})tan−1(x1+x+x2) + c

2 ^{-1}≤ft({{1+x+x²}{x}}) + c

Correct Answer is : 2 tan⁡−1(1+x+x2x)^{-1}≤ft({{1+x+x²}{x}})tan−1(x1+x+x2) + c

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AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 72 of 160

Marks: +1, -0

∫

\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}

dx =

Solution:

Consider the function

\int \frac{x-1}{(x+1)\sqrt{x^{3}+x^{2}+x}} \, dx

Simplify the expression

\int \frac{x^{2}-1}{(x+1)^{2}x\sqrt{1+\frac{1}{x}+x}} \, dx \;\; = \int \frac{x^{2}-1}{\left(1+1+\frac{1}{x}+x\right)x^{2}\sqrt{1+\frac{1}{x}+x}} \, dx

\;\; = \int \frac{1-\frac{1}{x^{2}}}{1+\left(1+\frac{1}{x}+x\right)\sqrt{1+\frac{1}{x}+x}} \, dx

Let

1+\frac{1}{x}+x = t^{2}

Then

\left(1-\frac{1}{x^{2}}\right) \, dx = 2t \, dt

This implies,

\int -\frac{2t \, dt}{(1+t^{2}) t} = 2 \int \frac{dt}{1+t^{2}}

= 2 \tan^{-1}(t) + c

= 2 \tan^{-1}\left(\sqrt{\frac{1+x+x^{2}}{x}}\right) + c

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