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

Lösung

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

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

Dezimale

- 1.80277 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

\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

Multipliziere aus

\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

Wende die Summenregel an:

\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))

Vereinfache

\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

d er i v a t i v e x^{4} (2 x - 5)^{6}

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