What is the integral of 6/((x+1)^2) ?

The integral of 6/((x+1)^2) is -6/(x+1)+C

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integral of 6/((x+1)^2)

\int \frac{6}{( x + 1 ) ^{2}} d x

- \frac{6}{x + 1} + C

Solution steps

\int \frac{6}{( x + 1 ) ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{( x + 1 ) ^{2}} d x

Apply u-substitution

= 6 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= 6 (- \frac{1}{u})

Substitute back

u = x + 1

= 6 (- \frac{1}{x + 1})

Simplify

6 (- \frac{1}{x + 1}) : - \frac{6}{x + 1}

= - \frac{6}{x + 1}

Add a constant to the solution

= - \frac{6}{x + 1} + C

x \to - 3 lim (\frac{3 - x}{x + 3})

\frac{\partial}{\partial y} (2 sec^{2} (2 x + 3 y))

\int 4 - 9 x^{2} d x

d er i v a t i v e e^{x - y}

d er i v a t i v e y = 7