Need help with some calculus integral problems?

Question

Integrate; indefinite: (x^4)(sqrt(x^5 + 1))dx

definite: 0(lower limit) to 1(upper limit): (x^2)/(x^3 + 5)dx

definite: e to 9: ((ln(x))/x)dx

indef: tan(x)dx "no formula"

indef: sqrt(x)e^(x(sqrt(x)))dx

indef: x(2x + 3)^3 dx

please list all steps

I have the first one and third one done

Answer 1

∫x^4 √(x^5 + 1) dx; let u = x^5 + 1, du = 5x^4 du
= ∫(1/5) u^(1/2) du
= (1/5) u^(3/2) / (3/2) + c
= (2/15) (x^5 + 1)^(3/2) + c.

∫(0 to 1) (x^2)/(x^3 + 5) dx; let u = x^3 + 5, du = 3x^2 dx
= ∫(5 to 6) (1/3) du / u
= (1/3) [ln u][5 to 6]
= (ln 6 - ln 5) / 3

∫(e to 9) ((ln x)/x) dx: let u = ln x, du = dx / x
= ∫(1 to ln 9) u du
= [u^2 / 2] [1 to ln 9] = ((ln 9)^2 - 1) / 2.

∫tan x dx
= ∫(sin x / cos x) dx; let u = cos x, du = -sin x dx
= ∫(-du / u)
= - ln |u| + c
= -ln |cos x| + c.

∫(√x e^(x√x)) dx; let u = x√x = x^(3/2), du = (3/2)√x dx
= ∫((2/3) e^u) du
= (2/3) e^u + c
= (2/3) e^(x√x) + c.

∫(x(2x + 3)^3 dx; let u = 2x+3, du = 2dx
= ∫((u-3)/2) u^3 (1/2) du
= (1/4) ∫(u^4 - 3u^3) du
= (1/4) (u^5 / 5 - 3u^4 / 4) + c
= (2x+3)^5 / 20 - 3(2x+3)^4 / 16 + c.

Answer 2

1: indefinite: (sqrt(x^5 + 1)) (x^4)dx
Substitution: u = x^5 du = 5x^4 dx, so x^4 dx = (1/5)du
So the integral is: Int sqrt(u + 1)(1/5)du = (2/15)(u+1)^(3/2) = (2/15)(x^5 + 1)^(3/2) +c

2: definite: 0(lower limit) to 1(upper limit): [1/(x^3 + 5)]x^2 dx

Substitution: u = x^3 du = 3x^2dx x^2 dx = (1/3)du

Int [1/(u + 5)] (1/3)du = (1/3) ln(u+5) = (1/3) ln(x^3 + 5)
Plugging in the limits gives:
(1/3)ln(6) - (1/3)ln(5) = (1/3) ln(6/5)

3: definite: e to 9: ((ln(x))/x)dx

Substitution: u = ln(x) du = (1/x)dx

Int ln(x)* (1/x)dx = Int u du = u^2/2 = (ln(x))^2/2)
Plugging in the limits:
(ln(9))^2/2 - (ln(e))^2/2 = (ln(9))^2/2 - 1/2

4: indef: tan(x)dx = Int sin(x)/cos(x) dx"no formula"

Substitution: u = cos(x) du = -sin(x) dx sin(x)dx = -du

Int -du/u = ln|u| = -ln|cos(x)| +c

5: indef: sqrt(x)e^(x(sqrt(x)))dx = Int x^(1/2) e^(x^(3/2))

Sub: u= x^(3/2) du = (3/2) x^(1/2)dx x^(1/2) dx = (2/3)du

Int e^(u) (2/3)du = (2/3) e^u = (2/3) e^(x^(3/2)) + c

6: indef: x(2x + 3)^3 dx

integration by parts will work:

dv = (2x + 3)^3 v= (1/8)(2x+3)^4 u = x du = 1
Int x(2x + 3)^3 = x * (1/8)(2x+3)^4 - Int[ (1/8)(2x+3)^4 ]

= (1/8) x (2x+3)^4 - (1/80) * (2x + 3)^5 + c

Answer 3

(x^4)(sqrt(x^5 + 1)dx
put x^5 + 1 = t
differentiating
5x^4 dx = dt
x^4 dx = dt/5
substituting
(1/5)sqrt(t) dt
integrating
1/5[t^(1/2 + 1)]/3/2
2/15[t^(3/2)]
2/15[x^5 + 1]^(3/2)

2)
[ln(x)/x]dx
[ln(x)(1/x)]dx
put ln(x) = y
differentiating
1/x dx = dy
substituting
y dy
integrating
y^2/2
1/2[ln(x)]^2 = 1/2[ln(9)^2 - ln(e)^2]
1/2[4 ln (3) - 2] = 2 ln(3) -1
3)
tan(x) dx
[sin(x)/cos(x)] dx
[d(-cos(x)/cosx]
integrating
- [ln(cosx)]

Answer 4

Integral(x^4)(sqrt(x^5 + 1))dx
Let t^2 =x^5+1 therefore 2tdt = 5x^4dx;
Substituting one get Integral(2/5)t^2dt = (2/15)t^3
= (2/15)(x^5 + 1)pow(3/2)

Integral (x^2)/(x^3 + 5)dx
let t=x^3+5 therefore dt=3x^2dx
Substituting we get Integtal (1/3)(1/t)dt = (1/3) log t
= (1/3) log (x^3 + 5);