What is the integral of 1/((x+4)^2) ?

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

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integral of 1/((x+4)^2)

integral

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

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

Solution steps

\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{36}{( x ^{4} - 4 x + 3 )}

\frac{1}{( ( x + 1 ) ( x - 2 ) )}

\frac{x}{( a + x ) ^{\frac{1}{2}}}

1 - cos (x) 1 + cos (x)

\frac{cos ^{3} ( x )}{sin ( x )}