maclaurin ln(1+4x)

Lösung

maclaurin

ln (1 + 4 x)

4 x - 8 x^{2} + \frac{64}{3} x^{3} - 64 x^{4} + \frac{1024}{5} x^{5} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{4}{1 !} x + \frac{- 16}{2 !} x^{2} + \frac{128}{3 !} x^{3} + \frac{- 1536}{4 !} x^{4} + \frac{24576}{5 !} x^{5} + \dots

Fasse zusammen

= 4 x - 8 x^{2} + \frac{64}{3} x^{3} - 64 x^{4} + \frac{1024}{5} x^{5} + \dots

\int \frac{3 x}{1 - x} d x

\int \frac{( x ^{2} - 4 ) ^{\frac{5}{2}}}{x} d x

x \to 0 lim (\frac{1}{2 x})

\int \frac{1}{( 5 - 4 x - x ^{2} ) ^{\frac{3}{2}}} d x

d er i v a t i v e y = (2 x)^{8}