# Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$

I recently saw someone using this formula here on one of the questions since I can't comment , can someone please give me the proof of this equation and type of problems where it can be used ?

• Lol , its actually pretty elementary only expand it and apply parts . (btw i was the one who referred the formula). – Sujith Sizon Oct 10 '15 at 3:19
• Looks like an application of product rule and chain rule – Triatticus Oct 10 '15 at 3:20

Open the brackets and separate it out as

$$\int e^{g(x)}g'(x)f(x)dx + \int e^{g(x)} f'(x)dx$$

Using integration by parts on the first terms we have

$$f(x)e^{g(x)} - \int f'(x)e^{g(x)} + \int f'(x)e^{g(x)}$$

The second and third term will cancell out yielding

$$=f(x)e^{g(x)}$$

Samples where it can be used :

Q1) $$\int e^{x+\frac{1}{x}}\left(1+x-\frac{1}{x}\right)dx$$

Q2) $$\int e^{\tan x}\left(x\sec^{2}x+ \sin2x\right)dx$$

And one of the beast

Q3) $$\int e^{x\sin x+\cos x}\left(\frac{x^{4}\cos^{3}x -x \sin x+ \cos x}{x^{2}\cos^{2}x}\right)dx$$

• Can you give me some example questions where this can be used ? – user270337 Oct 10 '15 at 3:22
• @VMiranda comment below if you aren't able to solve the third question . – Sujith Sizon Oct 10 '15 at 3:34

Hint: Take the derivative of the right side and use product rule.