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Popular >

integral of y/(1+y)

Solution

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

Solution

1 + y - ln ∣ 1 + y ∣ + C

Solution steps

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

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 = 1 + y

= 1 + y - ln ∣ 1 + y ∣

Add a constant to the solution

= 1 + y - ln ∣ 1 + y ∣ + C

Popular Examples

integral of sec(3x)

\int sec (3 x) d x

y^{^{'}} + 2 y = 7 x y^{2}

integral from 0 to 4 of (5t)/((t-5)^2)

\int_{0}^{4} \frac{5 t}{( t - 5 ) ^{2}} d t

integral of 2/((x-1)^3)

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

limit as x approaches 0+of 1/(e^x-1)-1/x

x \to 0 + lim (\frac{1}{e ^{x} - 1} - \frac{1}{x})

Frequently Asked Questions (FAQ)

What is the integral of y/(1+y) ?
The integral of y/(1+y) is 1+y-ln|1+y|+C