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

The integral of 1/((1+x/2)^2) is (4x)/((2+x)^2)+8(-1/(2+x)+1/((2+x)^2))+C

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

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

\frac{4 x}{( 2 + x ) ^{2}} + 8 (- \frac{1}{2 + x} + \frac{1}{( 2 + x ) ^{2}}) + C

Solution steps

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

Apply Integration By Parts

: \frac{4 x}{( 2 + x ) ^{2}} - \int - \frac{8 x}{( 2 + x ) ^{3}} d x

= \frac{4 x}{( 2 + x ) ^{2}} - \int - \frac{8 x}{( 2 + x ) ^{3}} d x

\int - \frac{8 x}{( 2 + x ) ^{3}} d x = - 8 (- \frac{1}{2 + x} + \frac{1}{( 2 + x ) ^{2}})

= \frac{4 x}{( 2 + x ) ^{2}} - (- 8 (- \frac{1}{2 + x} + \frac{1}{( 2 + x ) ^{2}}))

Simplify

= \frac{4 x}{( 2 + x ) ^{2}} + 8 (- \frac{1}{2 + x} + \frac{1}{( 2 + x ) ^{2}})

Add a constant to the solution

= \frac{4 x}{( 2 + x ) ^{2}} + 8 (- \frac{1}{2 + x} + \frac{1}{( 2 + x ) ^{2}}) + C

\frac{d}{d x} (x^{3} - 5 x)

\frac{d}{d x} ((1 - x) (x^{2} - 2)^{2})

x \to - 1 + lim (\frac{5}{x + 1})

y = csc (x) (x + cot (x))

f (x) = x^{5} e^{x}

What is the integral of 1/((1+x/2)^2) ?
The integral of 1/((1+x/2)^2) is (4x)/((2+x)^2)+8(-1/(2+x)+1/((2+x)^2))+C