integral from 1-5x to 1 of (u^3)/(1+u^2)

Lösung

\int_{1 - 5 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

\frac{1}{2} (- 25 x^{2} + 10 x - ln (2) + ln 25 x^{2} - 10 x + 2)

Schritte zur Lösung

\int_{1 - 5 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

Wende U-Substitution an

= \int_{25 x^{2} - 10 x + 2}^{2} \frac{v - 1}{2 v} d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{25 x^{2} - 10 x + 2}^{2} \frac{v - 1}{v} d v

Multipliziere aus

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \frac{1}{2} \cdot \int_{25 x^{2} - 10 x + 2}^{2} 1 - \frac{1}{v} d v

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_{25 x^{2} - 10 x + 2}^{2} 1 d v - \int_{25 x^{2} - 10 x + 2}^{2} \frac{1}{v} d v)

\int_{25 x^{2} - 10 x + 2}^{2} 1 d v = - 25 x^{2} + 10 x

\int_{25 x^{2} - 10 x + 2}^{2} \frac{1}{v} d v = ln (2) - ln 25 x^{2} - 10 x + 2

= \frac{1}{2} (- 25 x^{2} + 10 x - (ln (2) - ln 25 x^{2} - 10 x + 2))

Vereinfache

= \frac{1}{2} (- 25 x^{2} + 10 x - ln (2) + ln 25 x^{2} - 10 x + 2)

\int_{1}^{2} 5 x 5 - x^{2} d x

\int_{1}^{5} \frac{1}{1 + x - 1} d x

\int_{- 2}^{2} 1 - x^{2} d x

\int_{0}^{l n (2)} (e^{t} - e^{- t}) d t

\int_{1}^{6} 7 x + 1 d x