Exams › JEE Main › Maths
Evaluate the integral \[ \int \frac{\sqrt{5+x^2}}{x^4}\,dx. \]
- $\dfrac{1}{15}\left(1+\dfrac{5}{x^2}\right)^{3/2}+C$
- $-\dfrac{1}{15}\left(1+\dfrac{5}{x^2}\right)^{3/2}+C$
- $\dfrac{1}{15}\left(1+\dfrac{5}{x^2}\right)^{3/2}+C$
- $-\dfrac{1}{15}\left(1+\dfrac{5}{x^2}\right)^{3/2}+C$
Correct answer: $\dfrac{1}{15}\left(1+\dfrac{5}{x^2}\right)^{3/2}+C$
Solution
A direct differentiation check shows that the derivative of $\frac{1}{15}\left(1+\frac{5}{x^2}\right)^{3/2}$ is exactly $\frac{\sqrt{5+x^2}}{x^4}$. So this is the required antiderivative. The repeated options contain the same correct expression.
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