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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Integrate the function in Exercise:
$\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}$
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Answer

$\text{Let I }=\int\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx} \ \ \ \ \ ...\text{(i)}$
$\text{Let Linear}=\text{A}\frac{\text{d}}{\text{dx}}(\text{Quadratic})+\text{B}$
$\Rightarrow\ \ 5\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+4\text{x}+10\big)+\text{B}$
$\Rightarrow\ \ 5\text{x}+3=\text{A}(2\text{x}+4)+\text{B} \ \ \ \ ...\text{(ii)}$
$\Rightarrow\ \ 5\text{x}+3=2\text{A}\text{x}+4\text{A}+\text{B}$
Comparing coefficients of $x,$
$2\text{A}=5\ \ \Rightarrow\ \ \text{A}=\frac{5}{2}$
Comparing constants,
$4\text{A}+\text{B}=3$
On solving, we get
$\text{A}=\frac{5}{2}, \ \text{B}=-7$
Putting the values of $A$ and $B$ in eq. (ii),
$5\text{x}+3=\frac{5}{2}(2\text{x}+4)-7$
Putting this value of $5x + 3$ in eq. (i),
$\text{I}=\int\frac{\frac{5}{2}(2\text{x}+4)-7}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$\text{I}=\frac{5}{2}\int\frac{2\text{x}+4}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}-7\int\frac{1}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$\Rightarrow\ \ \text{I}=\frac{5}{2}\text{I}_1-7\ \text{I}_2\ \ \ \ ...\text{(iii)}$
$\text{Now I}_1=\int\frac{2\text{x}+4}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$\text{Putting }\text{ x}^2+4\text{x}+10=\text{t}\ \ \Rightarrow\ \ \ 2\text{x}+4=\frac{\text{dt}}{\text{dx}}\ \ \Rightarrow\ \ \ (2\text{x}+4)\text{ dx}=\text{dt}$
$\therefore\ \ \ \text{I}_1=\int\frac{\text{dt}}{\sqrt{\text{t}}}=\int\text{t}^{\frac{-1}{2}}\text{ dt}=\frac{\text{t}^{\frac{1}{2}}}{\frac{1}{2}}$
$2\sqrt{\text{t}}=2\sqrt{\text{x}^2+4\text{x}+10} \ \ \ \ ...\text{(iv)}$
$\text{Again I}_2=\int\frac{1}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$=\int\frac{1}{\sqrt{\text{x}^2+4\text{x}+4+6}}$
$=\int\frac{1}{\sqrt{(\text{x}+2)^2+\big(\sqrt{6}\big)^2}}\text{ dx}$
$=\log\begin{vmatrix}\text{x}+2+\sqrt{(\text{x}+2)^2+(6)^2}\end{vmatrix}$
$=\log\begin{vmatrix}\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\end{vmatrix} \ \ \ \ ...\text{(v)}$
Putting values of $I_1$ and $I_2$ in eq. $(iii),$
$\text{I}=5\sqrt{\text{x}^2+4\text{x}+10}-7\log\begin{vmatrix}\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\end{vmatrix}+\text{c}$

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