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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

f (x) = 3 x^{4} - 5 x^{3} - x^{2} + 4 x + 9

\int \frac{x}{x + 6} d x

\int \frac{4 cos ( x )}{sin ^{2} ( x )} d x

y = cos (x), at x = 1

\frac{d}{d x} (\frac{x ^{3} + 8}{x ^{2} - 4})

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