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

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

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

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

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

Solution steps

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

Simplify

x^{2} \frac{1}{x + 1} : \frac{x ^{2}}{x + 1}

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

Apply u-substitution

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

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

Expand

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

= \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

= \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{u ^{2}}{2} - 2 u + ln ∣ u ∣

Substitute back

u = x + 1

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

Add a constant to the solution

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

\int \frac{x y - 1}{x} d x

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

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

\int \frac{5}{x ^{2} + 6} d x

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