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

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

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

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

x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

= 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

= 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 ∣

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

Substitute back

u = 1 + x

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

Simplify

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

= x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3

Add a constant to the solution

= x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3 + C

\int x^{- 2} + 2 x - 6 x^{2} + cos (x) d x

\int (6 sinh (x) + x^{5}) d x

\int 6 x^{2} \cdot arccos (x) d x

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