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

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

Popular >

integral of 1/(y(1+y))

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

ln ∣ y ∣ - ln ∣ y + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{y ( 1 + y )} : \frac{1}{y} - \frac{1}{y + 1}

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

Apply the Sum Rule:

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

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

\int \frac{1}{y} d y = ln ∣ y ∣

\int \frac{1}{y + 1} d y = ln ∣ y + 1 ∣

= ln ∣ y ∣ - ln ∣ y + 1 ∣

Add a constant to the solution

= ln ∣ y ∣ - ln ∣ y + 1 ∣ + C

\int sin (\frac{1}{5} x) sin (\frac{1}{10} x) d x

\int 2 e^{2 x} x d x

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

\int \frac{1}{x ln ( 3 )} d x

\int \frac{10 s + 10}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

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