integral from 1 to t of 1/((2x+1)^3)

Lösung

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

\frac{t ^{2} + t - 2}{9 ( 2 t + 1 ) ^{2}}

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{3}^{1 + 2 t} \frac{1}{2 u ^{3}} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{3}^{1 + 2 t} \frac{1}{u ^{3}} d u

Wende die Potenzregel an

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{3}^{1 + 2 t}

Berechne die Grenzen:

- \frac{1}{2 ( 1 + 2 t ) ^{2}} + \frac{1}{18}

= \frac{1}{2} (- \frac{1}{2 ( 1 + 2 t ) ^{2}} + \frac{1}{18})

Vereinfache

= \frac{t ^{2} + t - 2}{9 ( 2 t + 1 ) ^{2}}

\int_{0}^{8} \frac{2 x}{x ^{2} + 1} d x

\int_{0}^{3} 4 r^{2} + 1 r d r

\int_{0}^{x} \frac{3}{8} x^{2} d x

\int_{0}^{2 π} 2 cos (t) sin (t) d t

\int_{0}^{2} r^{2} sin (θ) d r