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{e ^{x}}{1 + 36 e ^{2 x}} d x

\frac{5}{3} x^{3} - \frac{5}{2} x^{2}

\frac{4 x ^{2} + 4 x + 8}{x}

\int 3 cos (2 x) d x

f (x) = ln (2 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