Find the anti-derivative of...?

Question

Find the anti-derivative of...
1. (x+1)^2/x^2

2. intergral 1 to 8 of e^(x^1/3)/(x^2)^1/3

Answer 1

1. Integrate (x + 1)²/x² with respect to x.

First break apart the fraction. Then integrate.

(x + 1)²/x² = (x² + 2x + 1)/x² = 1 + 2/x + 1/x²

Now we can integrate.

∫{(x + 1)²/x²}dx = ∫{1 + 2/x + 1/x²}dx = x + 2ln|x| - 1/x + C
__________________

2. Find the integral from 1 to 8 of e^(x^⅓)/(x²)^⅓.

First rearrange the terms to make it easier to integrate.

e^(x^⅓)/(x²)^⅓ = e^(x^⅓)/(x^⅔) = (x^(-⅔))*e^(x^⅓)

Now we can integrate.

∫{e^(x^⅓)/(x²)^⅓}dx = ∫{(x^(-⅔))*e^(x^⅓)}dx
Let
u = x^⅓
du = ⅓x^(-⅔)dx
⅓du = x^(-⅔)dx

= ⅓∫{e^u}du = ⅓e^u = ⅓e^(x^⅓) [over 1 to 8]
= ⅓(e² - e) = ⅓e(e - 1)

Answer 2

1) The anti-derivative of f(x) is the function g(x) whose derivative w.r.t x, dg/dx, equals f(x)

2) g = x + 2ln(x) - ( 1 / x )

3) dg/dx = 1 + 2/x + 1/(x^2) =((x+1) / x)^2 = ((x + 1)^2) / x^2

Answer 3

Exponential applications are the main user-friendly to tell apart and combine as they do no longer substitute their sort whilst those operations are utilized to them. In different words, the respond to your 2d question: d(e^y)/dy = e^y it is so easy as that. A commonplace equation for it is as follows: If f(x) = e^kx then f '(x) = ke^kx that is by using chain rule (in case you go with for to be attentive to extra in this, i'm going to enable you be attentive to) meaning that with a nested function (you have the function kx interior) you desire to be useful to multiply by potential of its derivative besides. The opposite applies for integration: If f(x) = (a million/2)e^-2x then ?f(x)dx = -(a million/2)*(a million/2)e^-2x => ?f(x)dx = -(a million/4)e^-2x by using fact right here we would desire to divide by potential of the by-made of the interior function. wish the respond replaced into effective.

Answer 4

1.
(x+1)^2/x^2=(x^2+2x+1)/x^2=1+2/x+1/x^2=
int((1+2/x+1/x^2)dx)=
=int(1dx)+int(2/xdx)+int(1/x^2dx)=
=x+2*lnx-1/x+C

2.
e^(x^1/3)/(x^2)^1/3=
=e^(x^1/3)/x^2/3

u=x^1/3!
du=-1/3x^(-2/3)dx
du=-1/3/x^2/3dx

x=1, u=1^1/3=1
x=8, u=8^1/3=2

int(e^(x^1/3)/x^2/3dx)(1~8)=
=int(-3e^udu)(1~2)=
=-3e^u(1~2)=
=-3(e^2-e^1)=
=-3e(e-1)