English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x^4tan^3(x^5)

Lösung

\int x^{4} tan^{3} (x^{5}) d x

\frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5})) + C

Schritte zur Lösung

\int x^{4} tan^{3} (x^{5}) d x

Wende U-Substitution an

= \int \frac{3 u}{15 ( u ^{\frac{2}{3}} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{15} \cdot \int \frac{3 u}{u ^{\frac{2}{3}} + 1} d u

Wende U-Substitution an

= \frac{1}{15} \cdot \int \frac{3 v ^{3}}{v ^{2} + 1} d v

Entferne die Konstante:

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

= \frac{1}{15} \cdot 3 \cdot \int \frac{v ^{3}}{v ^{2} + 1} d v

Wende U-Substitution an

= \frac{1}{15} \cdot 3 \cdot \int \frac{w - 1}{2 w} d w

Entferne die Konstante:

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

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} \cdot \int \frac{w - 1}{w} d w

Multipliziere aus

\frac{w - 1}{w} : 1 - \frac{1}{w}

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} \cdot \int 1 - \frac{1}{w} d w

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}{15} \cdot 3 \cdot \frac{1}{2} (\int 1 d w - \int \frac{1}{w} d w)

\int 1 d w = w

\int \frac{1}{w} d w = ln ∣ w ∣

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} (w - ln ∣ w ∣)

Ersetze zurück

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} ((3 tan^{3} (x^{5}))^{2} + 1 - ln (3 tan^{3} (x^{5}))^{2} + 1)

Vereinfache

\frac{1}{15} \cdot 3 \cdot \frac{1}{2} ((3 tan^{3} (x^{5}))^{2} + 1 - ln (3 tan^{3} (x^{5}))^{2} + 1) : \frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5}))

= \frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5})) + C

\int sinh (5 x) d x

\int_{0}^{1} 8 cos (3 x) d x

f (x) = 4 cos^{2} (x)

\frac{\partial}{\partial x} (2 x y - x^{2} y)

\int 32 x^{3} - 15 x^{2} + 10 x d x