integrate $ \frac {(x^3 + 36)} {(x^2 + 36)}$

Question

I know I have to use long division first, but I don't really know how to do it in this case $$\int \frac{x^3 + 36}{x^2 + 36}dx$$

Answer 1

Hint: $x^3 + 36 = x(x^2+36) + 36(1-x)$. Now substitute $u = x^2+36$ to find $\int \frac{x}{x^2+36}dx$

Answer 2

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.

Answer 3

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 $$\int\frac{dx}{x^2+6^2}$$

Answer 4

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.

Answer 5

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.