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

The integral of y/(y^2+2y+1) is ln|y+1|+1/(y+1)+C

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integral of y/(y^2+2y+1)

\int \frac{y}{y ^{2} + 2 y + 1} d y

ln ∣ y + 1 ∣ + \frac{1}{y + 1} + C

Solution steps

\int \frac{y}{y ^{2} + 2 y + 1} d y

Complete the square

y^{2} + 2 y + 1 : (y + 1)^{2}

= \int \frac{y}{( y + 1 ) ^{2}} d y

Apply u-substitution

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

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

Expand

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

= \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

= \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}

= ln ∣ u ∣ - (- \frac{1}{u})

Substitute back

u = y + 1

= ln ∣ y + 1 ∣ - (- \frac{1}{y + 1})

Simplify

= ln ∣ y + 1 ∣ + \frac{1}{y + 1}

Add a constant to the solution

= ln ∣ y + 1 ∣ + \frac{1}{y + 1} + C

x \to 0 lim (6 x - 1)

\int \frac{- 3}{x ^{2} x ^{2} - 9} d x

\int cos (ln (7 x)) d x

x \to 0 lim (- x^{6})

x \to 4 - lim (\frac{x}{x + 4})

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