MHT CET · Maths · Indefinite Integration
\(\int \frac{1}{16 x^{2}+9} d x\) is equal to
- A \(\frac{1}{3} \tan ^{-1}\left(\frac{4 x}{3}\right)+c\)
- B \(\frac{1}{4} \tan ^{-1}\left(\frac{4 x}{3}\right)+c\)
- C \(\frac{1}{12} \tan ^{-1}\left(\frac{4 x}{3}\right)+c\)
- D \(\frac{1}{12} \tan ^{-1}\left(\frac{3 x}{4}\right)+c\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{12} \tan ^{-1}\left(\frac{4 x}{3}\right)+c\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \int \frac{1}{16 x^{2}+9} d x &=\frac{1}{16} \int \frac{1}{x^{2}+\left(\frac{3}{4}\right)^{2}} d x \\ &=\frac{1}{16} \times \frac{4}{3} \tan ^{-1}\left(\frac{x}{3 / 4}\right)+c \\ &=\frac{1}{12} \tan ^{-1}\left(\frac{4 x}{3}\right)+c \end{aligned}\)
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