How do I integrate:
∫ 2x/(x-4) dx
and
∫ (x+3)sqrt(x-2) dx
Thanks for help!
For the first one, make the substitution $\displaystyle u=x-4\implies x=u+4$ Thus,$\displaystyle \,du=\,dx$
The integral now becomes $\displaystyle \int\frac{2\left(u+4\right)}{u}\,du=\int\left(1+\f rac{8}{u}\right)\,du$. Can you take it from here?
For the second one, left $\displaystyle u=x-2\implies x=u+2$. Thus, $\displaystyle \,du=\,dx$
The integral now becomes $\displaystyle \int\left[\left[\left(u+2\right)+3\right]\sqrt{u}\right]\,du=\int\left(u^{\frac{3}{2}}+5u^{\frac{1}{2}}\ri ght)\,du$. Can you take it from here?
Does this make sense?
I would first tweek the integral before I apply integration by parts:
let $\displaystyle z=3x^4\implies x^4=\tfrac{1}{3}z$. Thus, $\displaystyle \,dz=12x^3\,dx\implies x^3\,dx=\tfrac{1}{12}\,dz$
Now, the integral becomes $\displaystyle \frac{1}{12}\int x^4\sin\left(z\right)\,dz=\frac{1}{36}\int z\sin\left(z\right)\,dz$
Now integration by parts will be a piece of cake from here.
Let $\displaystyle u=z$ and $\displaystyle \,dv=\sin\left(z\right)$
Why don't you take it from here and find $\displaystyle \,du$ and $\displaystyle v$.
PS: In the future, if you have new questions, please ask them in a new thread.
Here's something that may be helpful in the future:
To determine what $\displaystyle u$ should be, you can use the anagram LIPET. Go down the list (in this particular order) and see what you have in the integral [i.e. if you don't have logs, do you have inverse trig? etc]:
Logarithms (Natural) --------------Highest priority
Inverse Trigonometric Functions
Polynomial Expressions
Exponential Functions
Trigonometric Functions -----------Lowest Priority