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

The integral of (x^2-4)/(x+2) is ((x-2)^2)/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} + C

Solution steps

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

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

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

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

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

Substitute back

u = x - 2

= \frac{( x - 2 ) ^{2}}{2}

Add a constant to the solution

= \frac{( x - 2 ) ^{2}}{2} + C

\int - 2 x cos (2 x) d x

\int sin (3 q) + cos (3 q) d q

\int \frac{1 - 2 x}{16 - 9 x ^{2}} d x

\int x (3 x^{2} - 9)^{\frac{5}{2}} d x

\int \frac{x + 1}{x ^{2} - 3 x + 2} 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+C