{{x²+1}[(x²+1)-2 x]}{x^4}dx=

Correct Answer is : 19(1+1x2)3/2[2−3log⁡(1+1x2)]+c1/9 ≤ft(1+1/x²)³/2} ≤ft[2-3≤ft(1+1/x²)] + c91(1+x21)3/2[2−3log(1+x21)]+c

1/9 ≤ft(1+1/x²)³/2} ≤ft[2-3≤ft(1+1/x²)] + c

1/3 ≤ft(1+1/x²)¹/2} ≤ft[6- ≤ft(1+1/x²)²] + c

1/9 ≤ft(1+1/x²) ≤ft[3-2≤ft(1+1/x²)¹/2}] + c

1/3 ≤ft(1+1/x²)³/2} ≤ft[3- ≤ft(1+1/x²)] + c

Correct Answer is : 19(1+1x2)3/2[2−3log⁡(1+1x2)]+c1/9 ≤ft(1+1/x²)³/2} ≤ft[2-3≤ft(1+1/x²)] + c91(1+x21)3/2[2−3log(1+x21)]+c

Test Index

AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

\int \frac{\sqrt{x^2+1}[\log(x^2+1)-2\log x]}{x^4}dx=

$\frac{1}{9} \left(1+\frac{1}{x^2}\right)^{3/2} \left[2-3\log\left(1+\frac{1}{x^2}\right)\right] + c$
$\frac{1}{3} \left(1+\frac{1}{x^2}\right)^{1/2} \left[6- \log\left(1+\frac{1}{x^2}\right)^2\right] + c$
$\frac{1}{9} \left(1+\frac{1}{x^2}\right) \left[3-2\log\left(1+\frac{1}{x^2}\right)^{1/2}\right] + c$
$\frac{1}{3} \left(1+\frac{1}{x^2}\right)^{3/2} \left[3- \log\left(1+\frac{1}{x^2}\right)\right] + c$

Solution:

Consider the integral,

I=\int \frac{\sqrt{x^2+1}[\log(x^2+1)-2\log x]}{x^4}dx

\int \frac{1}{x^3} \sqrt{1+\frac{1}{x^2}} \log\left(1+\frac{1}{x^2}\right) dx

Let,

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

Then,

-2 \frac{dx}{x^3}=2t dt

So,

I=\int t^2 \log t^2 dt

=-2\int t^2 \log t dt

-2\left[ \frac{t^3}{3} \log t - \int \left( \frac{t^3}{3} \times \frac{1}{t} \right) dt \right]

=-\frac{2}{3} t^3 \log t + \frac{2}{9} t^3 + c

Further simplify the above,

\frac{1}{9} t^3 [2-3\log t^2] + c

= \frac{1}{9} \left(1+\frac{1}{x^2}\right)^{3/2} \left[2-3\log\left(1+\frac{1}{x^2}\right)\right] + c

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