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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} - 4 u + 8}{u} d u

= \int \frac{u ^{2} - 4 u + 8}{u} d u

Expand

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

= \int u - 4 + \frac{8}{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 4 d u + \int \frac{8}{u} d u

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

\int 4 d u = 4 u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

= \frac{u ^{2}}{2} - 4 u + 8 ln ∣ u ∣

Substitute back

u = x + 2

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

Add a constant to the solution

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

\int \frac{1}{sin ( x ) + 3 cos ( x ) + 1} d x

\frac{d}{d x} (sin (e^{x^{3} s i n (x)}))

\frac{d}{d x} (x^{2} - 8 x + 12)

\int \frac{1}{8 - 3 x} d x

\int_{0}^{1} - ln (x) d x

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