What is the integral of ((x^2+13x+42)/(x+6)) ?

The integral of ((x^2+13x+42)/(x+6)) is ((x+6)^2)/2+x+6+C

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integral of ((x^2+13x+42)/(x+6))

\int (\frac{x ^{2} + 13 x + 42}{x + 6}) d x

\frac{( x + 6 ) ^{2}}{2} + x + 6 + C

Solution steps

\int \frac{x ^{2} + 13 x + 42}{x + 6} d x

Apply u-substitution

: \int u + 1 d u

= \int u + 1 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 1 d u

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

\int 1 d u = u

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

Substitute back

u = x + 6

= \frac{( x + 6 ) ^{2}}{2} + x + 6

Add a constant to the solution

= \frac{( x + 6 ) ^{2}}{2} + x + 6 + C

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What is the integral of ((x^2+13x+42)/(x+6)) ?
The integral of ((x^2+13x+42)/(x+6)) is ((x+6)^2)/2+x+6+C