Question

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

Answer 1

$\int \frac{4x-1}{\sqrt{x^2+x-1}}dx=2\int \frac{2x+1}{\sqrt{x^2+x-1}}dx-\int \frac{3}{\sqrt{x^2+x-1}}dx$

$x^2+x-1=t(2x+1)dx=dt$

$=2\int \frac{dt}{\sqrt{t}}-\int \frac{3}{\sqrt{x^2+x-1}}dx$

$=4\sqrt{x^2+x-1}-\int \frac{3dx}{\sqrt{(x+\frac{1}{2}}{)}^2-\frac{5}{4}}$

$=4\sqrt{x^2+x-1}-\int \frac{3dx}{\sqrt{(x+\frac{1}{2}{)}^2-\frac{5}{4}}}$

put $x+\frac{1}{2}=tdx=dt$

$=4\sqrt{x^2+x-1}-3\log |t+\sqrt{t^2-(\frac{\sqrt{5}}{2}{)}^2}|+c$

$=4\sqrt{x^2+x-1}-3\log |x+\frac{1}{2}+\sqrt{(x+\frac{1}{2}{)}^2-\frac{5}{4}}|+c$