That “2x” is actually the derivative of x^2, if you notice.

When we differentiate e^something, it becomes e^something multiplied by the derivative of the something. Here, the something is x^2 and its derivative is 2x.

So, integrating 2xe^(x^2) with respect to x should get us

e^(x^2)

plus an arbitrary constant.

So is it true if ∫ f’(x)e^f(x) dx = e^f(x) + c

In general, yes; this is because differentiating e^f(x) gives us f’(x) e^f(x), so the reverse also holds.

Thanks!

Alternative : integration by substitution


Let u = x²
du/dx = 2x

du = 2x dx


∫ 2xe^(x²) dx

= ∫ e^(x²) 2x dx

= ∫ e^u du

= e^u + c

= e^(x²) + c


Of course, it is much easier to directly integrate since :

∫ f'(x) e^(f(x)) dx = e^((f(x))

Always looks out for directly integrable expressions, or if they can be made directly integrable

eg. ∫ cos 2x sin 2x dx

= ¼ ∫ 4 sin 2x cos 2x dx

= ¼ ∫ (2 sin 2x (2 cos 2x) ) dx

= ¼ sin² 2x + c ,


Since d/dx sin² 2x

= d/dx (sin 2x)²
= 2(sin 2x)(2cos 2x)
= 4 sin 2x cos 2x



Otherwise,


∫ cos 2x sin 2x dx

= ½ ∫ 2 sin 2x cos 2x dx

= ½ ∫ sin 4x dx

(Recall sin 2A = 2 sin A cos A. In this case your A = 2x and 2A = 4x)


= ½ (¼ (-cos 4x) + c
= -⅛ cos 4x + c

Both are valid.

Thank you!