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

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

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

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

\frac{( y + 1 ) ^{2}}{2} - 2 (y + 1) + 2 ln ∣ y + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \int u - 2 + \frac{2}{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 u d u - \int 2 d u + \int \frac{2}{u} d u

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

\int 2 d u = 2 u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + 2 ln ∣ u ∣

Substitute back

u = y + 1

= \frac{( y + 1 ) ^{2}}{2} - 2 (y + 1) + 2 ln ∣ y + 1 ∣

Add a constant to the solution

= \frac{( y + 1 ) ^{2}}{2} - 2 (y + 1) + 2 ln ∣ y + 1 ∣ + C

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + 1})

x 8 - x^{2}

\int \frac{1}{( 4 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial x} (x^{4} + x y^{3})

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

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