What is the integral of 1/(x^2+8x+16) ?

The integral of 1/(x^2+8x+16) is -1/(x+4)+C

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integral of 1/(x^2+8x+16)

\int \frac{1}{x ^{2} + 8 x + 16} d x

- \frac{1}{x + 4} + C

Solution steps

\int \frac{1}{x ^{2} + 8 x + 16} d x

Take the partial fraction of

\frac{1}{x ^{2} + 8 x + 16} : \frac{1}{( x + 4 ) ^{2}}

= \int \frac{1}{( x + 4 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x + 4

= - \frac{1}{x + 4}

Add a constant to the solution

= - \frac{1}{x + 4} + C

\frac{d}{d x} (lo g_{5} (e^{x}))

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f (x) = - 5 x^{3} - 7 x^{2} + 8 x - 7

What is the integral of 1/(x^2+8x+16) ?
The integral of 1/(x^2+8x+16) is -1/(x+4)+C