TS EAMCET · Maths · Indefinite Integration
\(\int \frac{d x}{x^2 \sqrt{4+x^2}}\) is equal to
- A \(\frac{1}{4} \sqrt{4+x^2}+C\)
- B \(\frac{-1}{4} \sqrt{4+x^2}+C\)
- C \(\frac{-1}{4 x} \sqrt{4+x^2}+C\)
- D \(\frac{9}{4 x} \sqrt{4+x^2}+C\)
Answer & Solution
Correct Answer
(C) \(\frac{-1}{4 x} \sqrt{4+x^2}+C\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Let } I=\int \frac{d x}{x^2 \sqrt{4+x^2}} \\ &=\int \frac{d x}{x^3 \sqrt{\frac{4}{x^2}+1}} \\ & \text { Put } \quad \frac{4}{x^2}+1=t \\ & \Rightarrow \quad-\frac{8}{x^3} d x=d t \\ & \therefore \quad I=\int \frac{d t}{-8 \sqrt{t}} \\ &=-\frac{1}{8}…
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