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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{2} (- \frac{1}{u})

Substitute back

u = 1 + x

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

Simplify

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

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

Add a constant to the solution

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

\frac{s}{( s ^{2} + 1 )}

y^{^{'}} = y cos (2 x), y (0) = 1.6

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

\int_{1}^{4} 5 - 2 t + 3 t^{2} d t

\frac{\partial}{\partial x} (2 x + 3 y + 4)

What is the integral of 1/(2(1+x)^2) ?
The integral of 1/(2(1+x)^2) is -1/(2(1+x))+C