integrate $ \frac {(x^3 + 36)} {(x^2 + 36)}$ I know I have to use long division first, but I don't really know how to do it in this case
$$\int \frac{x^3 + 36}{x^2 + 36}dx$$
 A: Hint: By inspection, we find that
$$x^3 + 36 = x (x^2 + 36)  - 36x + 36$$
so that
$$\frac{x^3 + 36}{x^2 + 36} = x + \frac{36 - 36x}{x^2 + 36}$$
Hence, the desired integral is equal to
$$\int x dx - 36 \int \frac{x - 1}{x^2 + 36} dx$$
To handle the right-most integral, break it up as
$$\int \frac{x}{x^2 + 36} dx - \int \frac{1}{x^2 + 6^2} dx$$
The first bit can be done with a substitution, and the second ought to look familiar.
A: The Maple code $$with(Student[Calculus1]): IntTutor((x^3+36)/(x^2+36), x);$$ produces it step by step with explanation. See that link and here for info.
A: Using http://en.wikipedia.org/wiki/Partial_fraction_decomposition#Procedure
$$ \frac{x^3+36}{x^2+36} = \frac{x(x^2+36)- 36x +36 }{x^2+ 36} $$
$$ \frac{x^3+36}{x^2+36} = \frac{x(x^2+36)}{x^2+36} - \frac {36x}{x^2+36} + \frac{36}{x^2+36} $$
$$ \int \frac{x^3+36}{x^2+36}dx  = \int \frac{x(x^2+36)}{x^2+36} dx -\int \frac {36x}{x^2+36}dx + \int\frac{36}{x^2+36}dx $$
$$ \int \frac{x^3+36}{x^2+36}dx = \int x*dx - 18\int\frac{2x}{x^2+36}dx + \int\frac{36}{x^2+36}dx $$
I hope now you can proceed.
A: Hint: $x^3 + 36 = x(x^2+36) + 36(1-x)$. Now substitute $u = x^2+36$ to find $\int \frac{x}{x^2+36}dx$
A: HINT:
Using Partial fraction decomposition $$\frac{x^3+36}{x^2+36}=x+A+B\frac{\frac{d(x^2+36)}{dx}}{x^2+36}+C\frac1{x^2+36}$$
$$\implies \frac{x^3+36}{x^2+36}=x+A+\frac{2Bx}{x^2+36}+\frac C{x^2+36}$$
$$\implies x^3+36=(x+A)(x^2+36)+2Bx+C$$
$$\implies x^3+36=x^3+Ax^2+x(36+2B)+36A+C$$
Compare the coefficients of the different power of $x$  to find $A,B,C$
Now, $$\int\frac{d f(x)}{dx}dx=\int d f(x)=f(x)+C$$ and we know about  $$\int\frac{dx}{x^2+6^2}$$
