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

The integral of (x^2)/(4+x) is ((4+x)^2)/2-8(4+x)+16ln|4+x|+C

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

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

\frac{( 4 + x ) ^{2}}{2} - 8 (4 + x) + 16 ln ∣ 4 + x ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{( u - 4 ) ^{2}}{u} d u

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

Expand

\frac{( u - 4 ) ^{2}}{u} : u - 8 + \frac{16}{u}

= \int u - 8 + \frac{16}{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 8 d u + \int \frac{16}{u} d u

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

\int 8 d u = 8 u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= \frac{u ^{2}}{2} - 8 u + 16 ln ∣ u ∣

Substitute back

u = 4 + x

= \frac{( 4 + x ) ^{2}}{2} - 8 (4 + x) + 16 ln ∣ 4 + x ∣

Add a constant to the solution

= \frac{( 4 + x ) ^{2}}{2} - 8 (4 + x) + 16 ln ∣ 4 + x ∣ + C

y^{^{'}} = - y^{\frac{3}{2}}

\int \frac{x ^{5} + 1}{x ^{6} - x ^{4}} d x

e^{- x} y^{^{'}} = \frac{x}{y}

x (x + 3) d y - y (2 x + 3) d x = 0

(x^{3} + 2)^{^{'}}

What is the integral of (x^2)/(4+x) ?
The integral of (x^2)/(4+x) is ((4+x)^2)/2-8(4+x)+16ln|4+x|+C