integral of ln(1+x^4)

Lösung

\int ln (1 + x^{4}) d x

x ln (1 + x^{4}) + \frac{1}{2} ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1) - \frac{1}{2} ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1) - 4 x + C

Schritte zur Lösung

\int ln (1 + x^{4}) d x

Wende die partielle Integration an

= x ln (1 + x^{4}) - \int \frac{4 x ^{4}}{1 + x ^{4}} d x

\int \frac{4 x ^{4}}{1 + x ^{4}} d x = - \frac{1}{2} ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1) + \frac{1}{2} ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1) + 4 x

= x ln (1 + x^{4}) - (- \frac{1}{2} ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1) + \frac{1}{2} ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1) + 4 x)

Vereinfache

= x ln (1 + x^{4}) + \frac{1}{2} ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1) - \frac{1}{2} ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1) - 4 x

Füge eine Konstante zur Lösung hinzu

= x ln (1 + x^{4}) + \frac{1}{2} ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1) - \frac{1}{2} ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1) - 4 x + C

n \to \infty lim (- 1)

\int_{1}^{2} - \frac{1}{16 z} + \frac{1}{4 z} d z

d er i v a t i v e 20 x (2 x^{2} + 7)^{4}

\frac{d}{d x} (- 5 x^{2} + 6 x)

\int_{0}^{π} 5 sin^{2} (2 x) d x