How do you integrate $\frac{x}{(ln 3) (x^{2} + 4)}$ $x/((ln3)(x^2 + 4))$ ?

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How do you integrate $\frac{x}{(ln 3) (x^{2} + 4)}$ $x/((ln3)(x^2 + 4))$ ?

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Answer 1 · 2015-05-02T08:08:29

$\int \frac{x}{(ln 3) (x^{2} + 4)} d x$ $intx/((ln3)(x^2+4))dx$

$ln 3$ $ln3$ is a constant so this turns down to,

$\Rightarrow \frac{1}{ln 3} \int \frac{x}{x^{2} + 4} d x = \frac{1}{ln 3} \int \frac{1}{2} \cdot \frac{2 x}{x^{2} + 4} d x = \frac{1}{2 ln 3} \int \frac{2 x}{x^{2} + 4} d x$ $=> 1/ln3intx/(x^2+4)dx =1/ln3 int1/2*(2x)/(x^2+4)dx=1/(2ln3)int(2x)/(x^2+4)dx$

The numerator is the derivative of the denominator so,

$\Rightarrow \frac{1}{2 ln 3} ln ∣ ∣ x^{2} + 4 ∣ ∣ + C$ $=>1/(2ln3)lnabs(x^2+4) + C$

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How do you integrate $\frac{x}{(ln 3) (x^{2} + 4)}$ $x/((ln3)(x^2 + 4))$ ?

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