What is the integral of (4x^3)/(x^2+x) ?

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

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integral of (4x^3)/(x^2+x)

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

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

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{x ^{3}}{x ^{2} + x} d x

Factor

\frac{x ^{3}}{x ^{2} + x} : \frac{x ^{2}}{x + 1}

= 4 \cdot \int \frac{x ^{2}}{x + 1} d x

Apply u-substitution

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

= 4 \cdot \int \frac{( u - 1 ) ^{2}}{u} d u

Expand

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

= 4 \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

= 4 (\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 ∣

= 4 (\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣)

Substitute back

u = x + 1

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

Add a constant to the solution

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

\frac{( x + 1 )}{x - 1}

y = sin (2 x) ln (x^{2}), (1, 0)

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

\int sinh (x) cosh (x) d x

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