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

The integral of (Inx)/((x+1)^2) is In(ln|1+x|+1/(1+x))+C

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

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

I n (ln ∣ 1 + x ∣ + \frac{1}{1 + x}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= I n (\int \frac{1}{u} d u - \int \frac{1}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

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

= I n (ln ∣ u ∣ - (- \frac{1}{u}))

Substitute back

u = 1 + x

= I n (ln ∣ 1 + x ∣ - (- \frac{1}{1 + x}))

Simplify

= I n (ln ∣ 1 + x ∣ + \frac{1}{1 + x})

Add a constant to the solution

= I n (ln ∣ 1 + x ∣ + \frac{1}{1 + x}) + C

What is the integral of (Inx)/((x+1)^2) ?
The integral of (Inx)/((x+1)^2) is In(ln|1+x|+1/(1+x))+C