integral from 1-9x to 1 of (u^3)/(1+u^2)

Lösung

\int_{1 - 9 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

\frac{1}{2} (- 81 x^{2} + 18 x - ln (2) + ln 81 x^{2} - 18 x + 2)

Schritte zur Lösung

\int_{1 - 9 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

Wende U-Substitution an

= \int_{81 x^{2} - 18 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_{81 x^{2} - 18 x + 2}^{2} \frac{v - 1}{v} d v

Multipliziere aus

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

= \frac{1}{2} \cdot \int_{81 x^{2} - 18 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_{81 x^{2} - 18 x + 2}^{2} 1 d v - \int_{81 x^{2} - 18 x + 2}^{2} \frac{1}{v} d v)

\int_{81 x^{2} - 18 x + 2}^{2} 1 d v = - 81 x^{2} + 18 x

\int_{81 x^{2} - 18 x + 2}^{2} \frac{1}{v} d v = ln (2) - ln 81 x^{2} - 18 x + 2

= \frac{1}{2} (- 81 x^{2} + 18 x - (ln (2) - ln 81 x^{2} - 18 x + 2))

Vereinfache

= \frac{1}{2} (- 81 x^{2} + 18 x - ln (2) + ln 81 x^{2} - 18 x + 2)

\int_{- 1}^{1} - x^{2} ∣ x ∣ d x

\int_{0}^{10} \frac{1}{10} x d x

\int_{- 1}^{1} (1 + x^{4} tan (x)) d x

\int_{0}^{π} x^{3} d x

\int_{- \infty}^{- 1} e^{- x} d x