Integral Calculus
86232
\(\int \frac{x^{2}}{(x+1)(x+2)^{2}} d x=\)
1 \(\log |\mathrm{x}+1|-\frac{4}{\mathrm{x}+2}+\frac{3}{(\mathrm{x}+2)^{2}}+\mathrm{c}\)
2 \(\log |\mathrm{x}+1|+\frac{1}{\mathrm{x}+2}+\mathrm{c}\)
3 \(\log |\mathrm{x}+1|-\frac{4}{\mathrm{x}+2}-\frac{3}{(\mathrm{x}+2)^{2}}+\mathrm{c}\)
4 \(\log |x+1|+\frac{4}{x+2}+c\)
Explanation:
(D) : \(I=\int \frac{x^{2}}{(x+1)(x+2)^{2}} d x\)
By partial fraction,
\(\frac{\mathrm{x}^{2}}{(\mathrm{x}+1)(\mathrm{x}+2)^{2}}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}+2}+\frac{\mathrm{C}}{(\mathrm{x}+2)^{2}}\)
\(\mathrm{x}^{2}=\mathrm{A}(\mathrm{x}+2)^{2}+\mathrm{B}(\mathrm{x}+1)(\mathrm{x}+2)+\mathrm{C}(\mathrm{x}+1)\)
When, \(\mathrm{x}=-2\),
\((-2)^{2}=\mathrm{C}(-2+1)\)
\(\mathrm{C}=-4\)
When, \(\mathrm{x}=-1\)
\((-1)^{2}=\mathrm{A}(2-1)^{2}\)
\(A=1\)
When, \(\mathrm{x}=0\)
\(0=\mathrm{A}(0+2)^{2}+\mathrm{B}(0+1)(0+2)+\mathrm{C}(0+1)\)
\(0=4 \mathrm{~A}^{2}+2 \mathrm{~B}+\mathrm{C}\)
\(-2 \mathrm{~B}=4 \times(1)^{2}=(-4)=4-4\)
\(\mathrm{B}=0\)
\(\frac{\mathrm{x}^{2}}{(\mathrm{x}+1)(\mathrm{x}+2)^{2}}=\frac{1}{\mathrm{x}+1}+\frac{-4}{(\mathrm{x}+2)^{2}}\)
\(\int \frac{x^{2}}{(x+1)(x+2)^{2}}=\int \frac{1}{x+1} d x+\int \frac{(-4)}{(x+2)^{2}} d x\)
\(=\log |\mathrm{x}+1|-4 \frac{(\mathrm{x}+2)^{(-2+1)}}{-2+1}+\mathrm{c}\)
\(=\log |\mathrm{x}+1|+\frac{4}{\mathrm{x}+2}+\mathrm{c}\)
