MHT CET · Maths · Indefinite Integration
The value of \(\int \frac{\mathrm{d} x}{7+6 x-x^2}\) is equal to
- A \(\frac{1}{4} \log \left(\frac{1+x}{7-x}\right)+\mathrm{c}\), (where c is a constant of integration)
- B \(\frac{1}{8} \log \left(\frac{7-x}{1+x}\right)+\mathrm{c},(\) where c is a constant of integration)
- C \(\frac{1}{4} \log \left(\frac{7-x}{1+x}\right)+\mathrm{c}\), (where c is a constant of integration)
- D \(\frac{1}{8} \log \left(\frac{1+x}{7-x}\right)+\mathrm{c}\), (where c is a constant of integration)
Answer & Solution
Correct Answer
(D) \(\frac{1}{8} \log \left(\frac{1+x}{7-x}\right)+\mathrm{c}\), (where c is a constant of integration)
Step-by-step Solution
Detailed explanation
\(\int \frac{1}{7+6 x-x^2} \mathrm{~d} x =\int \frac{1}{7+6 x-x^2-9+9} \mathrm{~d} x \)
\( =\int \frac{1}{16-\left(x^2-6 x+9\right)} \mathrm{d} x \)
\( =\int \frac{1}{4^2-(x-3)^2} \mathrm{~d} x \)
\( =\frac{1}{2(4)} \log \left|\frac{4+(x-3)}{4-(x-3)}\right|+\mathrm{c} \)
\( =\frac{1}{8} \log \left|\frac{1+x}{7-x}\right|+\mathrm{c}\)
\( =\int \frac{1}{16-\left(x^2-6 x+9\right)} \mathrm{d} x \)
\( =\int \frac{1}{4^2-(x-3)^2} \mathrm{~d} x \)
\( =\frac{1}{2(4)} \log \left|\frac{4+(x-3)}{4-(x-3)}\right|+\mathrm{c} \)
\( =\frac{1}{8} \log \left|\frac{1+x}{7-x}\right|+\mathrm{c}\)
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