integral of (t^2+t)e^{t^3}+3^t

Lösung

\int (t^{2} + t) e^{t^{3}} + 3^{t} d t

\frac{1}{3} e^{t^{3}} + \frac{1}{2} - \frac{t ^{2} Γ ( \frac{1}{\frac{3}{2}} , - ( t ^{2} ) ^{\frac{3}{2}} )}{\frac{3}{2} \frac{3}{2} - ( t ^{2} ) ^{\frac{3}{2}}} + \frac{3 ^{t}}{ln ( 3 )} + C

Schritte zur Lösung

\int (t^{2} + t) e^{t^{3}} + 3^{t} d t

Multipliziere aus

(t^{2} + t) e^{t^{3}} + 3^{t} : e^{t^{3}} t^{2} + e^{t^{3}} t + 3^{t}

= \int e^{t^{3}} t^{2} + e^{t^{3}} t + 3^{t} d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{t^{3}} t^{2} d t + \int e^{t^{3}} t d t + \int 3^{t} d t

\int e^{t^{3}} t^{2} d t = \frac{1}{3} e^{t^{3}}

\int e^{t^{3}} t d t = \frac{1}{2} - \frac{t ^{2} Γ ( \frac{1}{\frac{3}{2}} , - ( t ^{2} ) ^{\frac{3}{2}} )}{\frac{3}{2} \frac{3}{2} - ( t ^{2} ) ^{\frac{3}{2}}}

\int 3^{t} d t = \frac{3 ^{t}}{ln ( 3 )}

= \frac{1}{3} e^{t^{3}} + \frac{1}{2} - \frac{t ^{2} Γ ( \frac{1}{\frac{3}{2}} , - ( t ^{2} ) ^{\frac{3}{2}} )}{\frac{3}{2} \frac{3}{2} - ( t ^{2} ) ^{\frac{3}{2}}} + \frac{3 ^{t}}{ln ( 3 )}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} e^{t^{3}} + \frac{1}{2} - \frac{t ^{2} Γ ( \frac{1}{\frac{3}{2}} , - ( t ^{2} ) ^{\frac{3}{2}} )}{\frac{3}{2} \frac{3}{2} - ( t ^{2} ) ^{\frac{3}{2}}} + \frac{3 ^{t}}{ln ( 3 )} + C

x \to 0 lim (ln (x^{2}) (1 e^{x} + 1))

\frac{d}{d t} (3 cos^{3} (t))

l a pl a ce t r an s f or m 3^{(- \frac{1}{10} t)}

in v erse l a pl a ce \frac{- 5 e ^{(- 3 s)}}{( s ^{2} - 5 s + 6 )}

\int \frac{2 x ^{2} - 5}{x ^{4} - 5 x ^{2} + 6} d x