KCET · Maths · Indefinite Integration
If \(\int \frac{\sqrt{x}}{x(x+1)} d x=k \tan ^{-1} m\), then \((k, m)\) is
- A \((2, \mathrm{x})\)
- B \((1, \mathrm{x})\)
- C \((1, \sqrt{\mathrm{x}})\)
- D \((2, \sqrt{\mathrm{x}})\)
Answer & Solution
Correct Answer
(D) \((2, \sqrt{\mathrm{x}})\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
&\int \frac{\sqrt{x}}{x(x+1)} d x=k \tan ^{-1} m \\
&\text { Put } \quad\left\{\begin{array}{l}
x=\tan ^{2} \theta \\
d x=2 \tan \theta \cdot \sec ^{2} \theta d \theta
\end{array}\right. \\
&\quad=\int \frac{\tan \theta}{\tan ^{2} \theta \cdot \sec ^{2} \theta} \cdot\left(2 \tan \theta \cdot \sec ^{2} \theta\right) d \theta \\
&=2 \int d \theta=2 \theta=2 \tan ^{-1} \sqrt{x}=k \tan ^{-1}(m)
\end{aligned}\)
On comparing, we get \((\mathrm{k}, \mathrm{m})=(2, \sqrt{\mathrm{x}})\)
&\int \frac{\sqrt{x}}{x(x+1)} d x=k \tan ^{-1} m \\
&\text { Put } \quad\left\{\begin{array}{l}
x=\tan ^{2} \theta \\
d x=2 \tan \theta \cdot \sec ^{2} \theta d \theta
\end{array}\right. \\
&\quad=\int \frac{\tan \theta}{\tan ^{2} \theta \cdot \sec ^{2} \theta} \cdot\left(2 \tan \theta \cdot \sec ^{2} \theta\right) d \theta \\
&=2 \int d \theta=2 \theta=2 \tan ^{-1} \sqrt{x}=k \tan ^{-1}(m)
\end{aligned}\)
On comparing, we get \((\mathrm{k}, \mathrm{m})=(2, \sqrt{\mathrm{x}})\)
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