What is the integral of 1/((x+3)^2) ?

The integral of 1/((x+3)^2) is -1/(x+3)+C

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integral of 1/((x+3)^2)

integral

\frac{1}{( x + 3 ) ^{2}}

- \frac{1}{x + 3} + C

Solution steps

\int \frac{1}{( x + 3 ) ^{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 + 3

= - \frac{1}{x + 3}

Add a constant to the solution

= - \frac{1}{x + 3} + C

\frac{x}{( 1 + sin ^{2} ( x ) + cos ^{2} ( x ) )}

(x^{2} + 2)^{3}

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

5 e^{x}

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