What is the integral of (2x^2-3x+4)/(2x-7) ?

The integral of (2x^2-3x+4)/(2x-7) is (x^2)/2+2x+9ln|2x-7|+21/4+C

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integral of (2x^2-3x+4)/(2x-7)

\int \frac{2 x ^{2} - 3 x + 4}{2 x - 7} d x

\frac{x ^{2}}{2} + 2 x + 9 ln ∣ 2 x - 7 ∣ + \frac{21}{4} + C

Solution steps

\int \frac{2 x ^{2} - 3 x + 4}{2 x - 7} d x

Expand

\frac{2 x ^{2} - 3 x + 4}{2 x - 7} : \frac{2 x ^{2}}{2 x - 7} - \frac{3 x}{2 x - 7} + \frac{4}{2 x - 7}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 x ^{2}}{2 x - 7} d x - \int \frac{3 x}{2 x - 7} d x + \int \frac{4}{2 x - 7} d x

\int \frac{2 x ^{2}}{2 x - 7} d x = 2 (\frac{x ^{2}}{4} + \frac{7}{4} x + \frac{49}{8} ln ∣ 2 x - 7 ∣)

\int \frac{3 x}{2 x - 7} d x = \frac{3}{4} (2 x - 7 + 7 ln ∣ 2 x - 7 ∣)

\int \frac{4}{2 x - 7} d x = 2 ln ∣ 2 x - 7 ∣

= 2 (\frac{x ^{2}}{4} + \frac{7}{4} x + \frac{49}{8} ln ∣ 2 x - 7 ∣) - \frac{3}{4} (2 x - 7 + 7 ln ∣ 2 x - 7 ∣) + 2 ln ∣ 2 x - 7 ∣

Simplify

2 (\frac{x ^{2}}{4} + \frac{7}{4} x + \frac{49}{8} ln ∣ 2 x - 7 ∣) - \frac{3}{4} (2 x - 7 + 7 ln ∣ 2 x - 7 ∣) + 2 ln ∣ 2 x - 7 ∣ : \frac{x ^{2}}{2} + 2 x + 9 ln ∣ 2 x - 7 ∣ + \frac{21}{4}

= \frac{x ^{2}}{2} + 2 x + 9 ln ∣ 2 x - 7 ∣ + \frac{21}{4}

Add a constant to the solution

= \frac{x ^{2}}{2} + 2 x + 9 ln ∣ 2 x - 7 ∣ + \frac{21}{4} + C

\int \frac{x ^{6} - x ^{3} + 1}{x ^{4} + 9 x ^{2}} d x

x \to \infty lim (\frac{lo g _{4} ( x )}{lo g _{9} ( x )})

\int_{1}^{5} x - 1 d x

\int \frac{3}{7} x^{- \frac{1}{7}} d x

\frac{d}{d x} (\frac{x ^{2} - x}{1 + 3 x ^{2}})

What is the integral of (2x^2-3x+4)/(2x-7) ?
The integral of (2x^2-3x+4)/(2x-7) is (x^2)/2+2x+9ln|2x-7|+21/4+C