Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Integrate the function in Exercise: $(\text{x}^2+1)\text{log}\ \text{x}$
✓
Answer
Let $\text{I}=\int(\text{x}^2+1)\text{log x dx}=\int\text{x}^2\text{log x dx}+\int\text{log x dx}$
Let $I = I_1 + I_2....(1)$
where, $\text{I}_1=\int\text{x}^2\ \text{log x dx} \ \text{and I}_2=\int\text{logx}\ \text{dx} $
$\text{I}_1=\int\text{x}^2\text{log x dx}$
Taking $\log x$ as first function and $x^{2 }$ as secound function and integrating by parts,
we obtain $\int\text{I}.\text{II dx}=\text{I}\int\text{II dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\text{I}\int\text{II dx}\Big\}\text{dx}$
$\text{I}_1=\text{log x}\int\text{x}^2\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\text{log x}\Bigg)\int\text{x}^2\text{dx}\Bigg\}\text{dx}$
$=\text{log x}.\frac{\text{x}^3}{3}-\int\frac{1}{\text{x}}.\frac{\text{x}^3}{3}\text{dx}$
$=\frac{\text{x}^3}{3}\text{log x}-\frac{1}{3}\Big(\int\text{x}^2\text{dx}\Big)$
$=\frac{\text{x}^3}{3}\text{log x}-\frac{\text{x}^3}{9}+\text{C}\dots(2)$
$\text{I}_2=\int\text{log x dx}$
Taking $\log x$ as first function and $1$ as secound function and integrating by parts,
we obtain $\text{I}_2=\text{log x}\int1.\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\text{log x}\Bigg)\int1.\text{dx}\Bigg\}\text{dx}$
$=\text{log x}.\text{x}-\int\frac{1}{\text{x}}.\text{x dx}$
$=\text{x log x}-\int1\text{dx}$
$=\text{x log x}-\text{x}+\text{C}_2\dots(3)$
Using equations $(2)$ and $(3)$ in $(1),$ we obtain
$\text{I}=\frac{\text{x}^3}{3}\text{log x}-\frac{\text{x}^3}{9}+\text{C}_1+\text{x logx}-\text{x}+\text{C}_2$
$=\frac{\text{x}^3}{3}\text{log x}-\frac{\text{x}^3}{9}+\text{x logx}-\text{x}+(\text{C}_1+\text{C}_2)$
$=\Bigg(\frac{\text{x}^3}{3}+\text{x}\Bigg)\text{log x}-\frac{\text{x}^3}{9}-\text{x}+\text{C}$
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