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

The integral of (x^2)/(2(x+1)) is 1/4 (x^2-2x+2ln|x+1|-3)+C

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

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

\frac{1}{4} (x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{x ^{2}}{x + 1} d x

Apply u-substitution

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

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

Expand

\frac{( u - 1 ) ^{2}}{u} : u - 2 + \frac{1}{u}

= \frac{1}{2} \cdot \int u - 2 + \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

= \frac{1}{2} (\int u d u - \int 2 d u + \int \frac{1}{u} d u)

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = x + 1

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

Simplify

\frac{1}{2} (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣) : \frac{1}{4} (x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3)

= \frac{1}{4} (x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3)

Add a constant to the solution

= \frac{1}{4} (x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3) + C

\frac{d}{d x} (e^{9 x^{2}})

x^{^{''}} + 2 x = 0

y = x^{3} - 3 x + 1

\frac{\partial}{\partial x} (sin (x) + x^{2} e^{y} - 1)

f (x) = \frac{ln ( x )}{x ^{3}}

What is the integral of (x^2)/(2(x+1)) ?
The integral of (x^2)/(2(x+1)) is 1/4 (x^2-2x+2ln|x+1|-3)+C