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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals5 Marks
Question
Integrate the function $\frac{5 x+3}{\sqrt{x^{2}+4 x+10}}$
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Answer

Let $5x + 3 = A \frac{d}{d x}\left(x^{2}+4 x+10\right)+B$
$\Rightarrow 5x + 3 = A(2x + 4) + B$
Now, equating the coefficients of $x$ and constant term on both sides, we get,
$2A = 5$
$\Rightarrow A=\frac{5}{2}$
$4A + B = 3$
$\Rightarrow B = -7$
$\Rightarrow 5 x+3=\frac{5}{2}(2 x+4)-7$
Again, $\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{\frac{5}{2}(2 x+4)-7}{\sqrt{x^{2}+4 x+10}} d x$
$\Rightarrow \frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$
Now, let us consider, $\int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$
Let $x^2 + 4x + 10 = t$
$\Rightarrow (2x + 4) dx = dt$
$\therefore \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{dt}{\sqrt t}=2 \sqrt{t}=2 \sqrt{x^{2}+4 x+10} ......(i)$
And, Now let us consider, $\int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$
$\Rightarrow \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{1}{(\sqrt{x^{2}+4 x+4})+\sqrt6} d x$
$\Rightarrow \int \frac{1}{(x+2)^{2}+(\sqrt{6})^{2}} d x$
$=\log |(x+2) \sqrt{x^{2}+4 x+10}| .....(ii)$
using eq. $(i)$ and $(ii),$ we get,
$\Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\frac{5}{2}[2 \sqrt{x^{2}+4 x+10}]-7 \log (x+2) \sqrt{x^{2}+4 x+10} |+C$
$\Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=5 \sqrt{x^{2}+4 x+10}-7 \log |(x+2) \sqrt{x^{2}+4 x+10}|+C$

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