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)

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

- \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

y = 9 x^{2} ln (x), y = 36 ln (x)

cos^{2} (t)

x \to - 1 lim (x^{3} + x^{2} - 6 x + 9)

\int \frac{x}{( x + 2 ) ( 5 x - 1 )} d x

\frac{d ^{2}}{d x ^{2}} (tan (5 x + 2))