MHT CET · Maths · Indefinite Integration
\(\int \frac{x^{2}+1}{(x-3)(x-2)} d x=P x+Q \log |x-3|+R \log |x-2|+c\), where \(c\) is constant of
integration, then the values of \(\mathrm{P}, \mathrm{Q}, \mathrm{R}\) are, respectively
- A \(0,10,5\)
- B \(0,10,-5\)
- C \(1,10,5\)
- D \(1,10,-5\)
Answer & Solution
Correct Answer
(D) \(1,10,-5\)
Step-by-step Solution
Detailed explanation
(C)
\(I=\int \frac{x^{2}+1}{(x-3)(x-2)} d x\)
\(=\int \frac{\left(x^{2}-5 x+6\right)+(5 x-5)}{(x-3)(x-2)} d x\)
\(\quad=\int \frac{x^{2}-5 x+6}{x^{2}-5 x+6} d x+5 \int \frac{x-1}{(x-3)(x-2)} d x\)
Let \(\frac{(x-1)}{(x-3)(x-2)}=\frac{A}{(x-3)}+\frac{B}{(x-2)}\)
\(\therefore(x-1)=A(x-2)+B(x-3)\)
\(\therefore A+B=1\) and \(2 A+3 B=1 \Rightarrow A=2, B=-1\)
\(\therefore I=\int d x+5 \int\left[\frac{2}{(x-3)}-\frac{1}{(x-2)}\right] d x\)
Thus \(P=1, Q=10, R=-5\)
\(I=\int \frac{x^{2}+1}{(x-3)(x-2)} d x\)
\(=\int \frac{\left(x^{2}-5 x+6\right)+(5 x-5)}{(x-3)(x-2)} d x\)
\(\quad=\int \frac{x^{2}-5 x+6}{x^{2}-5 x+6} d x+5 \int \frac{x-1}{(x-3)(x-2)} d x\)
Let \(\frac{(x-1)}{(x-3)(x-2)}=\frac{A}{(x-3)}+\frac{B}{(x-2)}\)
\(\therefore(x-1)=A(x-2)+B(x-3)\)
\(\therefore A+B=1\) and \(2 A+3 B=1 \Rightarrow A=2, B=-1\)
\(\therefore I=\int d x+5 \int\left[\frac{2}{(x-3)}-\frac{1}{(x-2)}\right] d x\)
Thus \(P=1, Q=10, R=-5\)
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