Mathematics

# $\int \dfrac {4x-1}{\sqrt {x^{2}+x-1}}dx$

##### SOLUTION
$\int \dfrac{4x-1}{\sqrt{x^{2}+x-1}}dx = 2\int \dfrac{2x+1}{\sqrt{x^{2}+x-1}}dx - \int \frac{3}{\sqrt{x^{2}+x-1}}dx$
$x^{2}+x-1=t (2x+1)dx=dt$
$=2\int \dfrac{dt}{\sqrt{t}}-\int \dfrac{3}{\sqrt{x^{2}+x-1}}dx$
$=4\sqrt{x^{2}+x-1}-\int \dfrac{3 dx}{\sqrt{(x+\frac{1}{2}})^{2}-\dfrac{5}{4}}$
$= 4\sqrt{x^{2}+x-1}- \int \dfrac{3 dx}{\sqrt{(x+\frac{1}{2})^{2}-\dfrac{5}{4}}}$
put $x+\dfrac{1}{2}=t\, dx=dt$
$=4\sqrt{x^{2}+x-1}-3 \log |t+\sqrt{t^{2}-(\dfrac{\sqrt{5}}{2})^{2}}|+c$
$=4\sqrt{x^{2}+x-1}- 3 \log |x+\dfrac{1}{2}+\sqrt{(x+\frac{1}{2})^{2}-\dfrac{5}{4}}|+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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