What is the integral from-1 to 3 of (4x^2-7)/(2x+3) ?

The integral from-1 to 3 of (4x^2-7)/(2x+3) is (4ln(3)-8)/2

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integral from-1 to 3 of (4x^2-7)/(2x+3)

\int_{- 1}^{3} \frac{4 x ^{2} - 7}{2 x + 3} d x

\frac{4 ln ( 3 ) - 8}{2}

Decimal Notation

- 1.80277 \dots

Solution steps

\int_{- 1}^{3} \frac{4 x ^{2} - 7}{2 x + 3} d x

Apply u-substitution

: \int_{1}^{9} \frac{u ^{2} - 6 u + 2}{2 u} d u

= \int_{1}^{9} \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_{1}^{9} \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_{1}^{9} 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_{1}^{9} u d u - \int_{1}^{9} 6 d u + \int_{1}^{9} \frac{2}{u} d u)

\int_{1}^{9} u d u = 40

\int_{1}^{9} 6 d u = 48

\int_{1}^{9} \frac{2}{u} d u = 4 ln (3)

= \frac{1}{2} (40 - 48 + 4 ln (3))

Simplify

\frac{1}{2} (40 - 48 + 4 ln (3)) : \frac{4 ln ( 3 ) - 8}{2}

= \frac{4 ln ( 3 ) - 8}{2}

\frac{d}{d x} (I n)

\frac{d y}{d x} = 12 x^{3} y

\int (e^{x} + 1) (e^{x} + x) d x

x^{4} (2 x - 5)^{6}

y^{^{''}} + 40 y^{^{'}} + 625 y = 0

What is the integral from-1 to 3 of (4x^2-7)/(2x+3) ?
The integral from-1 to 3 of (4x^2-7)/(2x+3) is (4ln(3)-8)/2