Evaluate S 19e^(-t/5) dt?

Question

The S is the integral
This is part of my review for a test. I need to know how to do this step by step.

I'm going to need to ask a lot of questions today, so if you could also answer any of the questions below, I'd appreciate it.

Evaluate S (x^2+x+1)/(x^2+1) dx

or
Evaluate S (e^x)/ (e^x + 1)^(1/2) dx

or
Evaluate S (5x+16) / (x^2+9) dx

Answer 1

S 19e^(-t/5) dt

let u = -t/5
then du = -dt/5
and dt = -5du

the integral becomes

S 19e^(u)*(-5)du =
19*5 Se^(u)du =
95e^(u) + C=
95^(-t/5) + C


∫(e^x)/ (e^x + 1)^(1/2) dx
let e^x + 1 = u
then e^x dx = du

the integral becomes
∫du/(u)^1/2 =

∫u^(-1/2)du=

u^(1/2)/(1/2) + C=

2u^(1/2) + C=

2(e^x + 1)^1/2 + C


∫ (5x+16) / (x^2+9) dx=
∫5x/(x^2 + 9)dx + ∫16/(x^2 + 9)dx=

let x^2 + 9 = u
then 2xdx = du
and xdx = du/2

∫5*(du/2)/u + ∫16/(x^2 + 9)dx =

(5/2)ln(u) + 16∫1/(x^2 + 9)dx =

2.5 ln(x^2 + 9) + (16/3)arctan(x/3)

if you need to check some of your results just follow the link
http://integrals.wolfram.com/index.jsp

Answer 2

1) S 19e^(-t/5) dt = 19S e^(-t/5) dt (constant doesnt affect)
= 19/[1/ (-1/5) ] [ e^(-t/5) ] +C
= 19*-5 [[ e^(-t/5) ] +C
= -95e^(-t/5) + C
where C is a constant difference


(x^2+1) + x x
2) S (x^2+x+1)/(x^2+1) dx = S ________ dx = S 1 + __dx
x^2+1 x^2+1

= S [ 1 + 0.5 { 2x / (x^2+1) }dx = x + 0.5ln(x^2+1) +C

3) S (e^x)/ (e^x + 1)^(1/2) dx
let e^x = (tanθ)^2
if i differential this then it will give me
e^x = 2tanθ (secθ)^2
substitute back in original equation to give
= S [ 2tanθ (secθ)^2 ] / [ (tanθ)^2 +1 ]^0.5 dθ
= S [ 2tanθ (secθ)^2 ] / secθ dθ
since (tanθ)^2 + 1 =(secθ)^2
= S 2tanθ (secθ) dθ = 2 S tanθ (secθ) dθ
but d(secθ) = tanθ (secθ)
so S tanθ (secθ) = secθ
= 2 [ secθ ] + C


3) S (5x+16) / (x^2+9) dx = S 5x / (x^2+9) + 16/(x^2+9) dx
now lets concentrate on the first part first S 5x / (x^2+9) dx
integral of x^2 is 2x
that is S x^2 = 2x
so (5/2)* S (2x) / (x^2+9) dx = 2.5ln (x^2+9) + K

now second part S 16/(x^2+9) dx = 16 S 1/(x^2+9) dx
let x = 3tanθ
so dx = 3(secθ)^2 dθ
then by substitution
=16 S 3(secθ)^2 / [9(tanθ)^2+9] dθ
= 16*3 S (secθ)^2 / [9(secθ)^2] dθ
since (tanθ)^2 + 1 =(secθ)^2
=16*3/9 S dθ = 16θ/3

Answer 3

∫19e^(-t/5)dt =
let u = -t/5, du = (-1/5)dt, dt = -5du
∫-5*19e^udu =
-95e^u =
-95e^(-t/5) + C

∫(x^2+x+1)/(x^2+1) dx =
∫[(x^2 + 1)/(x^2 + 1)]dx + ∫xdx/(x^2 + 1) =
∫dx + (1/2)∫du/u = (using u = (x^2 + 1))
x + (1/2)ln(u) + C =
x + (1/2)ln(x^2 + 1) + C

∫(e^x)/ (e^x + 1)^(1/2) dx =
let u = (e^x + 1), du = e^xdx
∫u^(-1/2)du =
2u^(1/2) + C =
2(e^x + 1)^(1/2) + C

∫(5x + 16) / (x^2 + 9) dx =
∫ 5xdx/(x^2 + 9) + ∫16dx/(x^2 + 9) =
(5/2)Ln(x^2 + 9) + (16/3)tan^-1(x/3) + C