If \(f(x)\) is a polynomial of the second degree in \(x\) such that \(f(0)=f(1)=3, f(2)=-3\). Then, \(\int…
If \(f(x)\) is a polynomial of the second degree in \(x\) such that \(f(0)=f(1)=3, f(2)=-3\). Then, \(\int \frac{f(x)}{x^3-1} d x=\)
- \(\log \left(\frac{x^2+x+1}{|x-1|}\right)+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
- \(\log \left(\frac{x^2+x+1}{|x-1|}\right)-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
- \(\log \left(\frac{x^2+x+1}{|x-1|}\right)-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
- \(\log \left(\frac{x^2+x+1}{|x-1|}\right)+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
Solution
\(f(0)=f(1)=3 f(2)=-3\)
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,
\(\begin{aligned}
a & =1, b=-1 \\
\int \frac{f(x)}{x^3-1} d x & =\int \frac{x^2-x-3}{x^3-1} d x \\
\frac{x^2-x-3}{x^3-1} & =\frac{A}{(x-1)}+\left(\frac{B x+C}{x^2+x+1}\right) \\
& =\frac{x^2(A+B)+x(A-B+C)+(A-C)}{\left(x^3-1\right)}
\end{aligned}\)
On comparing \(A+B=1, A-B+C=-1, A-C=-3\)
\(\begin{aligned}
& \Rightarrow \quad A=-1, B=2, C=2 \\
& \quad \int\left(\frac{x^2-x-3}{x^3-1}\right) d x=\int \frac{-1}{(x-1)} d x+\int \frac{(2 x+2)}{\left(x^2+x+1\right)} d x \\
& =-\log |x-1|+\int \frac{(2 x+1) d x}{x^2+x+1}+\int \frac{1 \cdot d x}{x^2+x+1} d x \\
& =-\log \mid x-1)+\log \left(x^2+x+1\right)+\int \frac{d x-2)^2+\left(\sqrt{\frac{3}{2}}\right)^2}{(x+1} \\
& =\log \left(\frac{x^2+x+1}{|x-1|}\right)+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C
\end{aligned}\)
Asked in: AP EAMCET 2020 (17 Sep Shift 1)
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