calculus intergration?

Question

a) 3x^2 / x^3 - a^3 dx where a is a constant
b) 2x+4 / x^2+4x+1 dx
c) x squareroot 3x+4 dx

Answer 1

a) For ∫3x^2/(x^3 - a^3)dx, take x^3 - a^3 = u, then 3x^2dx =

du and ∫3x^2/(x^3 - a^3)dx = ∫du/u = lnu + C1 = ln(x^3 - a^3) +
C1.

b) For ∫(2x + 4)/(x^2 + 4x + 1)dx, take x^2 + 4x + 1 = v, then

(2x + 4)dx = dv and ∫(2x + 4)/(x^2 + 4x + 1)dx = ∫dv/v

= lnv + C2 = ln(x^2 + 4x + 1) + C2.


c) For ∫x√(3x + 4)dx, take (3x + 4) = t², so that x = (t² - 4)/3

and dx = (2/3)tdt. Therefore: ∫x√(3x + 4)dx

= ∫1/3(t² - 4)*t*(2/3)tdt = 2/9∫(t^4 - 4t²)dt

= 2/9[(t^5)/5 - 4(t^3)/3] + C3 = (2/135)t^3*(3t² - 20) + C3

= (2/135)(3x + 4)^(3/2)[3(3x + 4) - 20] + C3

= (2/135)(3x + 4)^(3/2)(9x - 8) + C3.

Answer 2

a) 3x^2 / (x^3 - a^3) dx where a is a constant
Let u = (x^3 - a^3), du = 3x^2 dx
∫ [ 3x^2 / (x^3 - a^3) ] dx = ln(x^3 - a^3) + C

b) 2x + 4 / (x^2 + 4x + 1) dx
Let u = (x^2 + 4x + 1), du = (2x + 4)dx
∫ [ 2x + 4 / (x^2 + 4x + 1) ] dx =
ln(x^2 + 4x + 1) + C

c) x squareroot 3x+4 dx

From Wolfram,
(2/135)(9x - 8)(3x + 4)^(3/2)

Answer 3

a) integ (3/x - a^3 )dx
= 3 ln x - a^3 x + c

b) integ(2x + 4 x^-2 + 4x + 1 ) dx
= x^2 - 4/ x + 2 x^2 + x + c

c) integ [ x sqr (3x + 4) ] dx . . . . using integration by parts
=2/9 x(3x +4)^1.5 - 4/405(3x+4)^2.5 + c

Answer 4

a) 3x^2 / x^3 - a^3 dx where a is a constant
3x^2=U' & x^3 - a^3=U
I 1/U du= Ln|U| + C
== Ln|x^3-a^3| + K