Question

integrate of 2x^3 / x^2-4?

Answer 1

\[\int\limits\frac{2x^{3}}{x^{2}-4} dx\]?

Answer 2

yup. :)

Answer 3

I have a correct answer but i dont have the solution. can you show me?

Answer 4

\[2\int\limits\frac{x^{3}}{x^{2}-4} dx\]
looks like a long division..

Answer 5

let x^2 -4 = t
thus dt = 2xdx
now 2x^3 dx= (2xdx)(x^2) = (t+4)dt
so we are left with
(t+4)/t dt


try now..

Answer 6

\[\int\limits {} \frac{x(x^{2}-4)+4x}{x^{2}-4}\]

Answer 7

\[2\int\limits x+\frac{4x}{x^{2}-4}\]

Answer 8

do you understand now???

Answer 9

\[x2+4\ln |x^{2} - 4|+C\]

Answer 10

that's the true answer.

Answer 11

\[2[\frac{x^{2}}{2}]\]
is the integral og the first term.....
for second term take
\[x^{2}-4=t\]
\[2xdx=dt\]

Answer 12

\[\int\limits \frac{4xdx}{x^{2}-4}=\int\limits \frac{2dt}{t}\]

Answer 13

\[=2\ln(t)\]

Answer 14

\[t=x^{2}-4\]
adding first and the second
\[2[\frac{x^{2}}{2}+2\ln|x^{2}-4|]+c\]
\[x^{2}+4\ln|x^{2}-4|+c\]