Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Evaluate the following integrals: $\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
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Answer
$\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)^2+2\text{x}\big(\text{x}^2+1\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)\big(\text{x}^2+1+2\text{x}\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{(\text{x}^2+1)(\text{x}+1)}\text{ dx}$
Let $\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{Cx}+\text{D}}{\text{x}^2+1}$
$\Rightarrow1=\text{A}(\text{x}+1)(\text{x}^2+1)+\text{B}(\text{x}^2+1)+(\text{Cx}+\text{D})(\text{x}+1)^2$
Putting $x = -1,$ we have
$1=2\text{B}$
$\Rightarrow\text{B}=\frac{1}{2}\ ...(\text{i})$
Putting $x = 0,$ we have
$\text{A}+\text{B}+\text{C}=1\ ...(\text{ii})$
Equating co$-$efficient of $x^3$ on both sides, we have
$\text{A}+\text{C}=0\ ...(\text{iii}) $
Equating co$-$efficient of $x^2$ on both sides, we have
$\text{A}+\text{B}+2\text{C}+\text{D}=0\ ...(\text{iv})$
$\Rightarrow2\text{C}=-1 [$Using $(i)]$
$\Rightarrow\text{C}=-\frac{1}{2}$
$\therefore\ \text{A}=\frac{1}{2} [$Using $(iii)]$
Putting $\text{A}=\frac{1}{2},\text{ B}=\frac{1}{2}$ and $\text{C}=-\frac{1}{2}$ in $(iv),$ we have
$\text{D}=0$
$\therefore\ \int_{0}^\limits{1}\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{\frac{1}{2}}{\text{x}+1}\text{ dx}+\int_{0}^\limits{1}\frac{\frac{1}{2}}{(\text{x}+1)^2}\text{ dx}+\int_{0}^\limits{1}\frac{-\frac{1}{2}\text{x}}{\text{x}^2+1}$
$=\Big[\frac{1}{2}\log(\text{x}+1)\Big]^1_0+\Big[\frac{1}{2}\times\Big(-\frac{1}{\text{x}+1}\Big)\Big]^1_0-\frac{1}{4}\int_{0}^\limits{1}\frac{2\text{x}}{\text{x}^2+1}\text{ dx}$
$=\frac{1}{2}\big(\log2-\log1\big)-\frac{1}{2}\Big(\frac{1}{2}-1\Big)-\Big[\frac{1}{4}\log(\text{x}^2+1)\Big]^1_0$
$=\frac{1}{2}\log2+\frac{1}{4}-\frac{1}{4}\big(\log2-\log1\big)$ $(\log1=0)$
$=\frac{1}{2}\log2+\frac{1}{4}\log\text{e}-\frac{1}{4}\log2$
$=\frac{1}{4}\log2+\frac{1}{4}\log_\text{e}$
$=\frac{1}{4}\big(\log2+\log_\text{e}\big)$
$=\frac{1}{4}\log(2\text{e})$
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