maclaurin ln(4+x^2)

Lösung

maclaurin

ln (4 + x^{2})

2 ln (2) + \frac{1}{4} x^{2} - \frac{1}{32} x^{4} + \frac{1}{192} x^{6} - \frac{1}{1024} x^{8} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

= 2 ln (2) + \frac{\frac{d}{d x} ( ln ( 4 + x ^{2} ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 4 + x ^{2} ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 4 + x ^{2} ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 2 ln (2) + \frac{0}{1 !} x + \frac{\frac{1}{2}}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- \frac{3}{4}}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{\frac{15}{4}}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{- \frac{315}{8}}{8 !} x^{8} + \dots

Fasse zusammen

= 2 ln (2) + \frac{1}{4} x^{2} - \frac{1}{32} x^{4} + \frac{1}{192} x^{6} - \frac{1}{1024} x^{8} + \dots

\int_{1}^{2} 15 4 x^{2} - 3 d x

x \to \infty lim (\frac{1}{8})

t an g e n t f (x) = \frac{3}{4 x + 1}, (- 1, - 1)

\frac{d}{d x} (2 x + \frac{200}{x})

d er i v a t i v e f (x) = \frac{32}{0}