AP EAMCET · Maths · Indefinite Integration
If \(f(x)\) is a polynomial of the second degree in \(x\) such that \(f(0)=f(1)=3, f(2)=-3\). Then, \(\int \frac{f(x)}{x^3-1} d x=\)
- A \(\log \left(\frac{x^2+x+1}{|x-1|}\right)+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
- B \(\log \left(\frac{x^2+x+1}{|x-1|}\right)-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
- C \(\log \left(\frac{x^2+x+1}{|x-1|}\right)-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
- D \(\log \left(\frac{x^2+x+1}{|x-1|}\right)+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
Answer & Solution
Correct Answer
(D) \(\log \left(\frac{x^2+x+1}{|x-1|}\right)+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
Step-by-step Solution
Detailed explanation
\(f(0)=f(1)=3 f(2)=-3\) Let \(f(x)=a x^2+b x+c\) \(\begin{aligned} & \Rightarrow \quad c=-3, a+b+c=-3 \Rightarrow(a+b=0) \\ & \text {and } \quad 4 a+2 b+c=-1 \Rightarrow 4 a+2 b=2 \end{aligned}\) and \(4 a+2 b+c=-1 \Rightarrow 4 a+2 b=2\) So,…
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