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

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

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integral of (x/(x+1))

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

x + 1 - ln ∣ x + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \int 1 - \frac{1}{u} 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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u - ln ∣ u ∣

Substitute back

u = x + 1

= x + 1 - ln ∣ x + 1 ∣

Add a constant to the solution

= x + 1 - ln ∣ x + 1 ∣ + C

\int tan^{3} (3 x) sec (3 x) d x

\int \frac{1}{( k - 1 ) ^{4}}

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

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

\frac{d ^{3}}{d x ^{3}} (4 x - 3)