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

The integral of (4x^2-7)/(2x+3) is 1/4 (4x^2-12x+4ln|2x+3|-27)+C

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

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

\frac{1}{4} (4 x^{2} - 12 x + 4 ln ∣ 2 x + 3 ∣ - 27) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} - 6 u + 2}{2 u} d u

= \int \frac{u ^{2} - 6 u + 2}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u ^{2} - 6 u + 2}{u} d u

Expand

\frac{u ^{2} - 6 u + 2}{u} : u - 6 + \frac{2}{u}

= \frac{1}{2} \cdot \int u - 6 + \frac{2}{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

= \frac{1}{2} (\int u d u - \int 6 d u + \int \frac{2}{u} d u)

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

\int 6 d u = 6 u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= \frac{1}{2} (\frac{u ^{2}}{2} - 6 u + 2 ln ∣ u ∣)

Substitute back

u = 2 x + 3

= \frac{1}{2} (\frac{( 2 x + 3 ) ^{2}}{2} - 6 (2 x + 3) + 2 ln ∣ 2 x + 3 ∣)

Simplify

\frac{1}{2} (\frac{( 2 x + 3 ) ^{2}}{2} - 6 (2 x + 3) + 2 ln ∣ 2 x + 3 ∣) : \frac{1}{4} (4 x^{2} - 12 x + 4 ln ∣ 2 x + 3 ∣ - 27)

= \frac{1}{4} (4 x^{2} - 12 x + 4 ln ∣ 2 x + 3 ∣ - 27)

Add a constant to the solution

= \frac{1}{4} (4 x^{2} - 12 x + 4 ln ∣ 2 x + 3 ∣ - 27) + C

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What is the integral of (4x^2-7)/(2x+3) ?
The integral of (4x^2-7)/(2x+3) is 1/4 (4x^2-12x+4ln|2x+3|-27)+C