integral of t^3ln(t^2)

Lösung

\int t^{3} ln (t^{2}) d t

\frac{1}{2} t^{4} ln (t) - \frac{t ^{4}}{8} + C

Schritte zur Lösung

\int t^{3} ln (t^{2}) d t

Wende die partielle Integration an

= \frac{1}{4} t^{4} ln (t^{2}) - \int \frac{t ^{3}}{2} d t

\int \frac{t ^{3}}{2} d t = \frac{t ^{4}}{8}

= \frac{1}{4} t^{4} ln (t^{2}) - \frac{t ^{4}}{8}

\frac{1}{4} t^{4} ln (t^{2}) = \frac{1}{2} t^{4} ln (t)

= \frac{1}{2} t^{4} ln (t) - \frac{t ^{4}}{8}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} t^{4} ln (t) - \frac{t ^{4}}{8} + C

e x p an d (- 2 cos (2 x) - sin (3 x) - 3)^{2}

y^{^{''}} + y = 0, y (0) = 0, y^{^{'}} (0) = 0

n = 1 \sum \infty \frac{n ^{n}}{n + 3}

\int \frac{sin ( 9 x )}{1 + cos ^{2} ( 9 x )} d x

d er i v a t i v e 1 - cos^{2} (x)