What is maclaurin ln^2(1+x) ?

The answer to maclaurin ln^2(1+x) is x^2-x^3+11/12 x^4-5/6 x^5+137/180 x^6+\ldots

maclaurin ln^2(1+x)

maclaurin

ln^{2} (1 + x)

x^{2} - x^{3} + \frac{11}{12} x^{4} - \frac{5}{6} x^{5} + \frac{137}{180} x^{6} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{2}{2 !} x^{2} + \frac{- 6}{3 !} x^{3} + \frac{22}{4 !} x^{4} + \frac{- 100}{5 !} x^{5} + \frac{548}{6 !} x^{6} + \dots

Refine

= x^{2} - x^{3} + \frac{11}{12} x^{4} - \frac{5}{6} x^{5} + \frac{137}{180} x^{6} + \dots

\int \frac{x ^{15}}{1 + x ^{16}} d x

\frac{d}{d x} ((\frac{x - 1}{x + 1})^{2})

\frac{d}{d x} ((6 x - 1) (3 x - 5)^{- 1})

t an g e n t f (x) = cos (\frac{π x}{2}), at x = 0.7

x y d x - (x^{2} - x) d y = 0

What is maclaurin ln^2(1+x) ?
The answer to maclaurin ln^2(1+x) is x^2-x^3+11/12 x^4-5/6 x^5+137/180 x^6+\ldots