^{-1}≤ft({2x+2}{{4x²+8x+13}})\,dx is equal to

Correct Answer is : (x+1)tan⁡−1(2x+23)−34log⁡(4x2+8x+139)+c(x+1)^{-1}≤ft(2x+2/3)-3/4≤ft({4x²+8x+13}{9})+c(x+1)tan−1(32x+2)−43log(94x2+8x+13)+c

(x+1)^{-1}≤ft(2x+2/3)-3/4≤ft({4x²+8x+13}{9})+c

3/2^{-1}≤ft(2x+2/3)-3/4≤ft({4x²+8x+13}{9})+c

(x+1)^{-1}≤ft(2x+2/3)-3/2≤ft(4x²+8x+13)+c

3/2(x+1)^{-1}≤ft(2x+2/3)-3/4≤ft(4x²+8x+13)+c

Correct Answer is : (x+1)tan⁡−1(2x+23)−34log⁡(4x2+8x+139)+c(x+1)^{-1}≤ft(2x+2/3)-3/4≤ft({4x²+8x+13}{9})+c(x+1)tan−1(32x+2)−43log(94x2+8x+13)+c

Test Index

VITEEE 2010 Paper

Question : 103 of 120

Marks: +1, -0

\int \sin^{-1}\left(\frac{2x+2}{\sqrt{4x^{2}+8x+13}}\right)\,dx

is equal to

$(x+1)\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log\left(\frac{4x^{2}+8x+13}{9}\right)+c$
$\frac{3}{2}\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log\left(\frac{4x^{2}+8x+13}{9}\right)+c$
$(x+1)\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{2}\log\left(4x^{2}+8x+13\right)+c$
$\frac{3}{2}(x+1)\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log\left(4x^{2}+8x+13\right)+c$

Solution:

Let

I=\int \sin^{-1}\left(\frac{2x+2}{\sqrt{4x^{2}+8x+13}}\right)\,dx

⇒

I=\int \sin^{-1}\left(\frac{2x+2}{\sqrt{4x^{2}+8x+4+9}}\right)\,dx

I=\int \sin^{-1}\left(\frac{2x+2}{\sqrt{(2x+2)^{2}+3^{2}}}\right)\,dx

Substituting 2x + 2 = 3 tan θ,

\Rightarrow 2\,dx = 3 \sec^{2} \theta \, d\theta

we get

I=\int \sin^{-1}\left(\frac{3 \tan \theta}{3 \sec \theta}\right) \cdot \frac{3}{2} \sec^{2} \theta \, d\theta

⇒

I=\frac{3}{2} \int \sin^{-1}(\sin\theta) \cdot \sec^{2} \theta \, d\theta

⇒

I=\frac{3}{2} \int \theta \sec^{2} \theta \, d\theta

\Rightarrow I=\frac{3}{2} \left[ \theta \tan \theta - \int \tan \theta \, d\theta \right] \quad \text{(integrating by parts)}

I=\frac{3}{2} \left[ \theta \tan \theta - \log |\sec \theta| \right] + c

=\frac{3}{2} \left[ \tan^{-1}\left(\frac{2x+2}{3}\right) \cdot \left(\frac{2x+2}{3}\right) - \log \sqrt{1+\left(\frac{2x+2}{3}\right)^{2}} \right] + c

=(x+1)\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log\left(\frac{4x^{2}+8x+13}{9}\right)+c

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