How to solve this integral? $I=\int{e^x \Big(2\log x+\frac{1}{x}+\frac{1}{x^2}\Big)dx}$

Question

$$I=\int{e^x \Big(2\log x+\frac{1}{x}+\frac{1}{x^2}\Big)dx}$$ I have tried to use many methods of substitution

But I am aware of a formula $$I=\int{e^x[f(x)+f'(x)]dx}=e^xf(x)+C$$

but can this formula be applied to the above equation

Answer 1

This is exactly how to solve the integral. Set $f(x):= 2 log(x) -\frac{1}{x}$

Then $f'(x)= \frac{2}{x}+\frac{1}{x^2}$

and therefore $f(x) + f'(x) = 2 log(x) +\frac{1}{x} +\frac{1}{x^2}$

We can then apply your formula and get $$I= \int e^x(2 log(x) +\frac{1}{x} +\frac{1}{2}) = e^x ( 2 log(x) -\frac{1}{x})$$

Answer 2

If you cannot see the what $f$ should be right away, here are some methods to get you in the right direction:

If we assume your factor in the integrand can be written as $$\log(x)+\frac{1}{x} +\frac{1}{x^2}=f(x)+f'(x)$$ you have a DE, with the inegrating factor of $e^x$, if you wanna solve it the "brute" way... alternatively you can fiddle around with some functions and see what fits the bill. Just remember that $\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}$ and $\frac{d}{dx}\log(x)=\frac{1}{x}$.

Answer 3

I always find it better to use more general form of integration by parts:

$$u= \ln x$$

$$v= e^x$$

$$\int u dv=uv-\int v du$$

$$2 \int e^x \ln x dx=2 e^x \ln x-2\int e^x \frac{1}{x} dx$$

The same way we get:

$$u= \frac{1}{x}$$

$$v= e^x$$

$$\int e^x \frac{1}{x} dx= e^x \frac{1}{x}+\int e^x \frac{1}{x^2} dx$$

Finally we have:

$$2 \int e^x \ln x dx+\int e^x \frac{1}{x} dx+\int e^x \frac{1}{x^2} dx=$$

$$=2 e^x \ln x -\int e^x \frac{1}{x} dx+\int e^x \frac{1}{x^2} dx=$$

$$=2 e^x \ln x -e^x \frac{1}{x}$$