Exercises using Imad Zak method which reduce the time without going in by parts

Exercise:

Integrate, $\int{\frac{\ln x}{{{x}^{2}}}dx}$ and $\int{\frac{\ln x}{\sqrt{x}}dx}$

Solution: Let $y=\frac{\ln x}{x}\Rightarrow y'=\frac{-1}{{{x}^{2}}}\ln x+\frac{1}{{{x}^{2}}}\Rightarrow \frac{\ln x}{{{x}^{2}}}=-y'+\frac{1}{{{x}^{2}}}$

So $\int{\frac{\ln x}{{{x}^{2}}}=\int{\left( -y'+\frac{1}{{{x}^{2}}} \right)dx}=-\int{y'\,dx}+\int{\frac{1}{{{x}^{2}}}dx}}=-\int{\frac{dy}{dx}dx}+\int{\frac{1}{{{x}^{2}}}dx}$

So $\int{\frac{\ln x}{{{x}^{2}}}dx}=-\int{dy+\int{\frac{1}{{{x}^{2}}}dx}=-y-\frac{1}{x}+c=-\frac{\ln x}{x}-\frac{1}{x}+c}$

Let $y=\sqrt{x}\ln x\Rightarrow y'=\frac{1}{2\sqrt{x}}\ln x+\frac{\sqrt{x}}{x}\Rightarrow \frac{\ln x}{2\sqrt{x}}=y'-\frac{\sqrt{x}}{x}\Rightarrow \frac{\ln x}{\sqrt{x}}=2y'-\frac{2\sqrt{x}}{x}$

So $\int{\frac{\ln x}{\sqrt{x}}dx}=2\int{y'}dx-2\int{\frac{\sqrt{x}}{x}dx}=2y-2\int{{{x}^{-1/2}}dx}=-2\sqrt{x}\ln x-4\sqrt{x}+c$ 

Imad Zak Method Technique for Integration :

Let $f$ be a map defined to be $f\left( x \right)=\frac{\ln x}{g\left( x \right)}\,$

where $g\left( x \right)\ne 0$ & $\deg \left( g\left( x \right) \right)=n\,\,,n\in \mathbb{R}$

Put $y\left( x \right)=\frac{\ln x}{h\left( x \right)}$ where $\deg \left( h\left( x \right) \right)=\deg \left( g\left( x \right) \right)-1=n-1$

Then $\int{\frac{\ln x}{g\left( x \right)}dx}=\int{\frac{dy}{dx}dx}+\int{\frac{d\left( \ln x \right)}{dx}h\left( x \right)dx}$