How do you integrate $\int 4 x {(x^{2} + 3)}^{- 3} d x$ $int4x(x^2+3)^(-3) dx$ ?

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Answer 1 · 2015-10-18T22:32:09

$- \frac{1}{{(x^{2} + 3)}^{2}} + C$ $-1/(x^2+3)^2 + C$

$\int 4 x {(x^{2} + 3)}^{- 3} d x = 2 \int {(x^{2} + 3)}^{- 3} 2 x d x =$ $int4x(x^2+3)^(-3) dx = 2 int (x^2+3)^(-3) 2xdx =$

$= 2 int (x^2+3)^(-3) (x^2+3)'dx = 2 int (x^2+3)^(-3) d(x^2+3) =$

$= 2 (x^2+3)^-2/-2 +C = -1/(x^2+3)^2 + C$

How do you integrate int4x(x^2+3)^(-3) dx∫4x(x2+3)−3dxint4x(x^2+3)^(-3) dx?