What is the integral of u/((1+u)) ?

The integral of u/((1+u)) is 1+u-ln|1+u|+C

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integral of u/((1+u))

integral

\frac{u}{( 1 + u )}

1 + u - ln ∣ 1 + u ∣ + C

Solution steps

\int \frac{u}{1 + u} d u

Apply u-substitution

: \int \frac{v - 1}{v} d v

= \int \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \int 1 - \frac{1}{v} d v

Apply the Sum Rule:

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

= \int 1 d v - \int \frac{1}{v} d v

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= v - ln ∣ v ∣

Substitute back

v = 1 + u

= 1 + u - ln ∣ 1 + u ∣

Add a constant to the solution

= 1 + u - ln ∣ 1 + u ∣ + C

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

\frac{x ^{4}}{4}

15 x^{2} (5 x^{3} - 6)^{5}

\frac{x ^{2}}{x ^{2} - 4}

\frac{1}{( x ^{3} - x )}