Integration functions

Question

I am unsure how to solve this problem:

$$\frac{2}{3}\int \frac{x^4 \;dx}{x^2+3} $$

Could anybody help me ?

Answer 1

HINT :

$I = \displaystyle\frac{2}{3}\int \frac{x^4 \;dx}{x^2+3}$

Performing polynomial long division gives you ;

$ I = \displaystyle\frac23\int\bigg(\frac9{x^2+3}+x^2-3\bigg)\,dx$

I assume you can do the rest

Answer 2

First divide $x^4 $ by $ x^2+3$ to get $$\frac{2}{3}\int \frac{x^4 \;dx}{x^2+3}= \frac{2}{3}\int [(x^2-3)+\frac{9 }{x^2+3}]dx$$

Now you have two integrals that you can solve .

Answer 3

Hint $$I=\frac{2}{3}\int \frac{x^4 \;dx}{x^2+3}=\frac{2}{3}\int \frac{x^4 -9+9\;}{x^2+3}dx$$ $$I=\frac{2}{3}\int \frac{(x^2-3)(x^2+3)+9}{x^2+3}dx$$ $$I=\frac{2}{3}\int {(x^2-3)}dx+6\int \frac{dx}{x^2+3}$$

Use arctan function for the last integral..