Finding indefinite Integrals ?

Question

I have a big Calculus midterm 2mrw afternoon and i can't seem to get these problems right, and i'm gonna be tested on them...if anyone can help, it would be greatly appreciated.

I wrote out the problems and put it in an image, so its easier to see:

http://img.photobucket.com/albums/v509/RiceRocketX/math.jpg

Answer 1

1)  Let u = 1 + x^3.  Then, du = 3x^2 dx, so

∫ (x^2)/(1+x^3)^2 = 1/3 ∫ du/u^2 = 1/3*[-1/(u)] + C = -1/(3 + 3x^3) + C

2) Let u = x^2 - 1.  Then, du = 2x dx, so

∫ x(x^2-1)^7 dx = 1/2 ∫ u^7 du = 1/2*[(u^8)/8] + C = (u^8)/16 + C = [(x^2-1)^8]/16 + C

Answer 2

The trick to this is you have to find some part of the numerator that is the derivative of another part of the function...

#1

In this case we say...

u = 1 + x^3

Therefore du = 3x^2

We have x^2 in the numerator, so all we need to do is multiply by 3 and also divide by 3 in the front of the integral....

INT ( x^2 / (1 + x^3)^2 ) dx

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

= (1/3) INT ( 1 / u^2)  du = (1/3)( - 1/u) + C

= -(1/3)( 1 / (1 + x^3) ) + C  ANSWER

#2

We'll say u = x^2 - 1

du = 2x dx

We already have x as part of the integral, so we just multiply it by 2 and also divide by 2 (or multiply by 1/2) out front of the integra...

INT ( x(x^2 - 1)^7 ) dx

= (1/2) INT ( 2x(x^2 - 1)^7 ) dx

= (1/2) INT ( u^7 ) du

= (1/2) (1/8) u^8 + C

= (1/16) (x^2 -1)^8 + C  ANSWER

Hope that helps.  The trick is seeing that one part of the integral is the derivative of another.  Then you use that derivative as du.

Take care,

David

Answer 3

1) int(x^2/(1+x^3)^2 dx)

take u=1+x^3

du=3x^2dx

du/3=x^2dx

int(du/3u^2)

(1/3)int(du/u^2)

(1/3)int(u^-2 du)

(1/3) (u^-1/-1)

=-1/3u

=-1/(3(1+x^3)) +c

2) int(x(x^2 -1)^7 dx)

take u=x^2-1

du=2x dx

du/2=x dx

now substitute

int(u^7 du/2)

=0.5int(u^7 du)

=0.5u^8/8

=u^8/16

=((x^2-1)^8)/16 +c

Answer 4

1.)

∫[x²/(1 + x^3)²]dx

u = 1 + x^3

du/dx = 3x²

du = 3x²dx

1/3du = x²dx

= ∫[1/u²]1/3du

= 1/3∫[1/u²]du

= 1/3∫[u^(-2)]du

= 1/3[-u^(-1)] + C

= -1/u + C

= -1/(1 + x^3) + C

2.)

∫[x(x² - 1)^7]dx

u = x² - 1

du/dx = 2x

du = 2xdx

1/2du = xdx

= ∫[u^7]1/2du

= 1/2∫[u^7]du

= 1/2[(u^8)/8] + C

= (u^8)/4 + C

= [(x² - 1)^8]/4 + C