AP EAMCET · Maths · Indefinite Integration
\(\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=\)
- A \(x \sin ^{-1} x+\sqrt{1-x^2}+c\)
- B \(\sin ^{-1} x+\sqrt{1-x^2}+c\)
- C \(x \sin ^{-1} x+c\)
- D \(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+c\)
Answer & Solution
Correct Answer
(C) \(x \sin ^{-1} x+c\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Given, } \int \frac{\sqrt{1-x^2} \cdot \sin ^{-1} x+x}{\sqrt{1-x^2}} d x \\ & \int\left(\frac{\sqrt{1-x^2} \cdot \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}\right) d x \\ & \int\left(\sin ^{-1} x+\frac{x}{\sqrt{1-x^2}}\right) d x \\ & \int \sin…
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