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
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