Mathematics

# Evaluate:$\int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}$.

##### SOLUTION
$I = \int \dfrac{5x+3}{\sqrt{x^{2}+4x+10}}dx=5\int \dfrac{x+\dfrac{3}{5}}{\sqrt{x^{2}+4x+10}}dx$
$=\dfrac{5}{2} \int \dfrac{2x+\dfrac{6}{5}}{\sqrt{x^{2}+4x+10}}dx$
$=\dfrac{5}{2} \int \dfrac{2x+4-4+\dfrac{6}{5}}{\sqrt{x^{2}+4x+10}}dx$
$=\dfrac{5}{2}\int \dfrac{2x+4}{\sqrt{x^{2}+4x+10}}dx+\dfrac{5}{2}\int \dfrac{\dfrac{-14}{5}}{\sqrt{x^{2}+4x+10}}dx$
$=\dfrac{5}{2} \int \dfrac{2x+4}{\sqrt{x^{2}+4x+10}}dx - 7\int \dfrac{dx}{\sqrt{x^{2}+4x+10}}dx$

Let $I_{1}=\dfrac{5}{2} \int \dfrac{2x+4}{\sqrt{x^{2}+4x+10}}dx$ and $I_{2}=7\int \dfrac{dx}{\sqrt{x^{2}+4x+10}}dx$

$I_{1}=\dfrac{5}{2} \int \dfrac{2x+4}{\sqrt{x^{2}+4x+10}}dx$
Let $x^{2}+4x+10=t$
$(2x+4)dx=dt$
Thus, $I_{1}=\dfrac{5}{2} \int \dfrac{1}{\sqrt{t}}dt$

$I_{1}=\dfrac{5}{2} \dfrac{t^{\dfrac{1}{2}}}{\dfrac{1}{2}}+C_{1}$

$I_{1}=5\sqrt{t}+C_{1}$
$I_{1}=5\sqrt{x^{2}+4x+10}+C_{1}$

Now, $I_{2}=7\int \dfrac{dx}{\sqrt{x^{2}+4x+10}}dx$

$I_{2}=7\int \dfrac{dx}{\sqrt{(x+2)^{2}+(\sqrt{6})^{2}}}dx$

$I_{2} =7 [log|x+2+\sqrt{x^{2}+4x+10}|]+C_{2}$

Therefore, $I=5\sqrt{x^{2}+4x+10}-7log|x+2+\sqrt{x^{2}+4x+10}|+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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