integral from 1 to 5 of 5/(x^2+4x+4)

Lösung

\int_{1}^{5} \frac{5}{x ^{2} + 4 x + 4} d x

\frac{20}{21}

Dezimale

0.95238 \dots

Schritte zur Lösung

\int_{1}^{5} \frac{5}{x ^{2} + 4 x + 4} d x

Entferne die Konstante:

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

= 5 \cdot \int_{1}^{5} \frac{1}{x ^{2} + 4 x + 4} d x

Vervollständige das Quadrat

x^{2} + 4 x + 4 : (x + 2)^{2}

= 5 \cdot \int_{1}^{5} \frac{1}{( x + 2 ) ^{2}} d x

Wende U-Substitution an

= 5 \cdot \int_{3}^{7} \frac{1}{u ^{2}} d u

Wende die Potenzregel an

= 5 [- \frac{1}{u}]_{3}^{7}

Berechne die Grenzen:

\frac{4}{21}

= 5 \cdot \frac{4}{21}

Vereinfache

= \frac{20}{21}

\frac{\partial}{\partial y} (\frac{x ^{2}}{1 + y ^{3}})

\frac{d}{d x} (a x e^{b x^{2}})

\frac{\partial}{\partial x} (3 x^{2} + 6 x y)

\frac{d}{d x} (\frac{12 ( 3 x ^{3} - 16 )}{4 x - x ^{4}})

\int p (p + 7)^{8} d p