What is the integral from-2 to 2 of 1/((x+4)(x-4)) ?

The integral from-2 to 2 of 1/((x+4)(x-4)) is -1/4 ln(3)

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integral from-2 to 2 of 1/((x+4)(x-4))

\int_{- 2}^{2} \frac{1}{( x + 4 ) ( x - 4 )} d x

- \frac{1}{4} ln (3)

Decimal

- 0.27465 \dots

Solution steps

\int_{- 2}^{2} \frac{1}{( x + 4 ) ( x - 4 )} d x

Take the partial fraction of

\frac{1}{( x + 4 ) ( x - 4 )} : - \frac{1}{8 ( x + 4 )} + \frac{1}{8 ( x - 4 )}

= \int_{- 2}^{2} - \frac{1}{8 ( x + 4 )} + \frac{1}{8 ( x - 4 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 2}^{2} \frac{1}{8 ( x + 4 )} d x + \int_{- 2}^{2} \frac{1}{8 ( x - 4 )} d x

\int_{- 2}^{2} \frac{1}{8 ( x + 4 )} d x = \frac{1}{8} ln (3)

\int_{- 2}^{2} \frac{1}{8 ( x - 4 )} d x = - \frac{1}{8} ln (3)

= - \frac{1}{8} ln (3) - \frac{1}{8} ln (3)

Simplify

- \frac{1}{8} ln (3) - \frac{1}{8} ln (3) : - \frac{1}{4} ln (3)

= - \frac{1}{4} ln (3)

i = 0 \sum \infty 2^{i}

y^{^{'}} = \frac{y ^{2}}{x ^{2}} + \frac{y}{x}

\frac{x ^{2} + 4}{3} + \frac{11 - x}{4}

\int_{0}^{2} 5 x^{2} d x

6 sin (3 x)

What is the integral from-2 to 2 of 1/((x+4)(x-4)) ?
The integral from-2 to 2 of 1/((x+4)(x-4)) is -1/4 ln(3)