Grade 12Integral Calculus

evaluate the integral of “sqrt(ln(x))” with respect to “x” .

jaswanth

10 Years agoGrade 12

1 Answer

jagdish singh singh

Approved Tutor Answer10 Years ago

$\hspace{-0.6 cm} $Let $\bf{I=\int \sqrt{\ln (x)}dx\;,}$ Now Put $\bf{\ln(x)=t^2\Rightarrow x=e^{t^2}\;,}$\\\\ Then $\bf{dx=2te^{t^2}dt.\;\; }$So we get $\bf{\int t\left( e^{t^2}2t\right)dt = te^{t^2}-\int e^{t^2}dt}$\\\\So $\bf{I=te^{t^2}-\int \left[1+\frac{t^2}{1}+\frac{t^4}{2!}+\frac{t^6}{3!}+.........\infty\right] = te^{t^2}-\left[t+\frac{t^3}{1}+\right]}$\\\\ So we get $\bf{I = x\cdot \sqrt{\ln x}-\left[(\ln x)^{\frac{1}{2}}+\frac{(\ln x)^{\frac{3}{2}}}{2!}+\frac{(\ln x)^{\frac{5}{2}}}{3!}+......\infty\right]+\mathcal{C}}$$