indefinite integral

[jon12]

I am working on a review sheet and having trouble with indefinite integrals. If anyone could help me out with showing me how to do these i would appreciate it.

(1+3u)^5

z^2 SquareRoot(1+z^3) dz

(4-x) squareroot(x) dx

(-4t)/(t^2+2)^6 dt

(8z^2+4z-18)/(z^4) dz

(sinu)/(cos^6u) du

(X^2-1)/(squareroot2x-1) dx

 

[galactus]

I am working on a review sheet and having trouble with indefinite integrals. If anyone could help me out with showing me how to do these i would appreciate it.

Here are some hints on a few to get you started.

\(\displaystyle \int(1+3u)^{5}\)

EDIT: easier way than expanding. Let \(\displaystyle w=1+3u, \;\ \frac{dw}{3}=du\)

\(\displaystyle \int z^{2}\sqrt{1+z^{3}} dz\)

Make a u substitution. Let \(\displaystyle u=1+z^{3}, \;\ \frac{du}{3}=z^{2}dz\)

\(\displaystyle \int (4-x)\sqrt{x} dx\)

Rewrite as \(\displaystyle 4\int \sqrt{x}dx-\int x^{\frac{3}{2}}dx\).

\(\displaystyle \int\frac{sin(x)}{cos^{6}(x)}dx\)

Let \(\displaystyle u=cos(x), \;\ -du=sin(x)dx\)

 

[jon12]

galactus

i been working on the indefinite integral (8z^2+4z-18)/(z^4) dz
worked it down to 8z^-1/-1 + 4z^-2/-2 - 18z^-3/-3

cannot figure out where to go from there

 

[galactus]

When you simplify this, it becomes:

\(\displaystyle 8\int z^{-2}dz+4\int z^{-3}dz-18\int z^{-4}dz\)

Now, it is an easy integration. Opposite of differentiation.

Remember, add 1 to the exponent and divide into the coefficient.

The first one would be \(\displaystyle -8z^{-1}=\frac{-8}{z}\)

 

[Deleted member 4993]

galactus

i been working on the indefinite integral (8z^2+4z-18)/(z^4) dz
worked it down to 8z^-1/-1 + 4z^-2/-2 - 18z^-3/-3

cannot figure out where to go from there

You have it correct - perhaps needs a little simplification.

\(\displaystyle \int \left (\frac{8z^2 \ + \ 4z \ - 18}{z^4}\right )dz \ = \ -\frac{8}{z} \ - \ \frac{2}{z^2} \ + \ \frac{6}{z^3} \ + \ C\)

Another way to express the same result would be:

\(\displaystyle \frac{2}{z^3}(3 \ - \ 4z)(z \ + \ 1) \ + \ C\)