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

Lösung

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

\frac{1}{15}

Dezimale

0.06666 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende die Potenzregel an

= \frac{1}{2} [- \frac{1}{u}]_{3}^{5}

Berechne die Grenzen:

\frac{2}{15}

= \frac{1}{2} \cdot \frac{2}{15}

Vereinfache

= \frac{1}{15}

\int_{5}^{7} x^{2} + 2 x + 2 - 18 d x

\int_{- 2}^{2} (10 x^{9} + 25 x^{3}) d x

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

\int_{1}^{y} \frac{1}{y ^{2}} d y

\int_{0}^{2} \frac{1}{1 + x ^{3}} d x