Calculus integration problem?

Question

How would you integrate: 5x / (2x^2+1)^1/2
plz explain

Answer 1

∫5x/√(2x²+1) dx
Let u=2x²+1, du=4x dx
∫5/(4√u) du
5/4 ∫u^(-1/2) du
-5/2 u^(1/2) + C
-5/2 √(2x²+1) + C

Answer 2

I see the x^2 in the denominator and 5x in the numerator. The derivative of x^2 is 2x, which can easily be made from 5x. You can do a "u - sub" if you're just starting integrals.

let u = 2x^2+1
du = 4xdx
So 1/u^1/2* du=4x / (2x^2+1)^1/2 dx, but we want 5/4's of that. so

integral 5x / (2x^2+1)^1/2 = integral (5/4) * u^-1/2 du
=5/2 u ^ (1/2)

Sub back in for u
= 5/2 (2x^2 + 1) ^ (1/2) + C

Answer 3

No, I believe you guys are forgetting the antiderivative of 1/x is ln x


So yeah you do the u-sub for 2x^2 +1
du = 4x

so 5x/sqrt U du/4x

5/4 integral 1/sqrt u

5/4 ln u + C

then you sub it back in
5/4 ln (sqrt 2x^2 + 1) + C

Answer 4

hi bill ok, the two the solutions arrived at are one and the comparable ya. the two would desire to be simplified as ? e^x sin x dx = [e^x (sin x - cos x)] / 2 ?sin(x)e^x dx= -(e^x)cos(x) + [(e^x)(sin(x)) - ?sin(x)e^x dx] After simplification, it is going to become as you wrote 2?sin(x)e^x dx= -(e^x)cos(x) + (e^x)(sin(x)) extra suitable simplified into 2?sin(x)e^x dx= (e^x)[(sin(x) - cos x)] So ? e^x sin x dx = [e^x (sin x - cos x)] / 2 surprising. you have written the stairs so of course. yet in basic terms made a slip interior the previous couple of steps. Congrats and all the wonderful ya.