What is the integral of (x^2-3x+7)/(x+5) ?

The integral of (x^2-3x+7)/(x+5) is ((x+5)^2)/2-13(x+5)+47ln|x+5|+C

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integral of (x^2-3x+7)/(x+5)

\int \frac{x ^{2} - 3 x + 7}{x + 5} d x

\frac{( x + 5 ) ^{2}}{2} - 13 (x + 5) + 47 ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{x ^{2} - 3 x + 7}{x + 5} d x

Apply u-substitution

: \int \frac{u ^{2} - 13 u + 47}{u} d u

= \int \frac{u ^{2} - 13 u + 47}{u} d u

Expand

\frac{u ^{2} - 13 u + 47}{u} : u - 13 + \frac{47}{u}

= \int u - 13 + \frac{47}{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 13 d u + \int \frac{47}{u} d u

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

\int 13 d u = 13 u

\int \frac{47}{u} d u = 47 ln ∣ u ∣

= \frac{u ^{2}}{2} - 13 u + 47 ln ∣ u ∣

Substitute back

u = x + 5

= \frac{( x + 5 ) ^{2}}{2} - 13 (x + 5) + 47 ln ∣ x + 5 ∣

Add a constant to the solution

= \frac{( x + 5 ) ^{2}}{2} - 13 (x + 5) + 47 ln ∣ x + 5 ∣ + C

lo g_{2} (x^{2} + 1)

f (x) = \frac{x}{x + 5}

y = 9 - 4 x, at x = - 1

\int_{0}^{\frac{π}{4}} 9 tan^{3} (x) d x

\frac{d}{d x} (\frac{5 x ^{4} + 3 x ^{3}}{4 x - 9})

What is the integral of (x^2-3x+7)/(x+5) ?
The integral of (x^2-3x+7)/(x+5) is ((x+5)^2)/2-13(x+5)+47ln|x+5|+C