MHT CET · Maths · Indefinite Integration
The value of \(\int \frac{d x}{x\left(x^{n}+1\right)}\) is
- A \(\frac{1}{n} \log \left(\frac{x^{n}}{x^{n}+1}\right)+C\)
- B \(\log \left(\frac{x^{n}+1}{x^{n}}\right)+C\)
- C \(\frac{1}{n} \log \left(\frac{x^{n}+1}{x^{n}}\right)+C\)
- D \(\log \left(\frac{x^{n}}{x^{n}+1}\right)+C\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{n} \log \left(\frac{x^{n}}{x^{n}+1}\right)+C\)
Step-by-step Solution
Detailed explanation
Let
\(I=\int \frac{d x}{x\left(x^{n}+1\right)} \)
\( \left.\text { (let } t=x^{n}+1, d t=n x^{n-1} d x\right) \)
\(=\int \frac{d t}{r x^{n} \cdot t} \quad\left(\frac{d t}{n x^{n}}=\frac{d x}{x}\right) \)
\(=\frac{1}{n} \int \frac{d t}{t(t-1)} \)
\(=\frac{1}{n} \int\left\{\frac{1}{t-1}-\frac{1}{t}\right\} d t \)
\(=\frac{1}{n}\{\log (t-1)-\log t\}+C \)
\(=\frac{1}{n} \log \frac{t-1}{t}+C \)
\(=\frac{1}{n} \log \frac{x^{n}}{x^{n}+1}+C \)
\(I=\int \frac{d x}{x\left(x^{n}+1\right)} \)
\( \left.\text { (let } t=x^{n}+1, d t=n x^{n-1} d x\right) \)
\(=\int \frac{d t}{r x^{n} \cdot t} \quad\left(\frac{d t}{n x^{n}}=\frac{d x}{x}\right) \)
\(=\frac{1}{n} \int \frac{d t}{t(t-1)} \)
\(=\frac{1}{n} \int\left\{\frac{1}{t-1}-\frac{1}{t}\right\} d t \)
\(=\frac{1}{n}\{\log (t-1)-\log t\}+C \)
\(=\frac{1}{n} \log \frac{t-1}{t}+C \)
\(=\frac{1}{n} \log \frac{x^{n}}{x^{n}+1}+C \)
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