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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

= 3 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 3 (- \frac{1}{u})

Substitute back

u = x + 1

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

Simplify

3 (- \frac{1}{x + 1}) : - \frac{3}{x + 1}

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

Add a constant to the solution

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

\int 1 + \frac{9}{4} x d x

\int 1 + x^{3} d x

f (x) = \frac{x ^{2}}{x - 3}

e^{2 x} cos (3 x)

f (x) = 4 tan (x), at x = \frac{π}{3}