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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}+1}{(\text{x}^2+1)(\text{x}+2)}\text{ dx}$
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Answer

Let $\text{I}=\int\frac{\text{x}^2+\text{x}+1}{(\text{x}^2+1)(\text{x}+2)}\text{ dx}$
We express
$\frac{\text{x}^2+\text{x}+1}{(\text{x}^2+1)(\text{x}+2)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\text{x}^2+\text{x}+1=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+2)$
Equating the coefficient of $x^2$​​​​​​​, x and constant, we get
$1=\text{A}+\text{B}\text{ and }1=2\text{B}+\text{C}\text{ and }1=\text{A}+2\text{C}$
$\text{or}\text{ A } =\frac{3}{5}\text{ and }\text{B}=\frac{2}{5}\text{ and }\text{C}=\frac{1}{5}$
$ \therefore\text{I}=\int\bigg(\frac{\frac{3}{5}}{\text{x}+2}+\frac{\frac{2}{5}\text{x}+\frac{1}{5}}{\text{x}^2+1}\bigg)\text{dx}$
$=\frac{3}{5}\int\frac{1}{\text{x}+2}\ \text{dx}+\frac{2}{5}\int\frac{\text{x}}{\text{x}+1}\ \text{dx}+\frac{1}{5}\int\frac{1}{\text{x}^2+1}\ \text{dx}$
$=\frac{3}{5}\text{I}_1+\frac{2}{5}\text{I}_2+\frac{1}{5}\text{I}_3\ \dots(1)$
Let x + 2 = u
On Differentiating both sides, we get
$\text{dx}=\text{du}$
$\text{I}_1=\int\frac{1}{\text{u}}\text{du}$
$=\log|\text{u}|+\text{C}_1$
$=\log|\text{x}+2|+\text{C}_1\ \dots(2)$
And, $\text{I}_2=\int\frac{\text{x}}{\text{x}+1}\text{ dx}$
Let $(\text{x}^2+1)=\text{u}$
On differentiating both sides, we get
$2\text{x}\ \text{dx}=\text{du}$
$\therefore\text{I}_2=\frac{1}{2}\int\frac{1}{\text{u}}\text{ du}$
$=\frac{1}{2}\log|\text{u}|+\text{C}_2$
$=\frac{1}{2}\log|\text{x}^2+1|+\text{C}_2\ \dots(3)$
And $\text{I}_3=\int\frac{1}{\text{x}^2+1}\ \text{dx}$
$=\tan^{-1}\text{x}+\text{C}_3\ \dots(4)$
From (1), (2), (3) and (4) we get
$\therefore\text{I}=\frac{3}{5}(\log|\text{x}+2|+\text{C}_1)+\frac{2}{5}(\frac{1}{2}\log|\text{x}^2+1|+\text{C}_2)\\+\frac{1}{5}\tan^{-1}\text{x}+\text{C}$
$=\frac{3}{5}\log|\text{x}+2|+\frac{1}{5}\log|\text{x}^2+1|+\frac{1}{5}\tan^{-1}\text{x}+\text{C}$
Hence, $\int\frac{\text{x}^2+\text{x}+1}{(\text{x}^2+1)(\text{x}+2)}\ \text{dx}=\frac{3}{5}\log|\text{x}+2|+\frac{1}{5}\tan^{-1}\text{x}+\text{C}$

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