What is maclaurin ln(1/(x+5))+xcos(x) ?

The answer to maclaurin ln(1/(x+5))+xcos(x) is -ln(5)+4/5 x+1/50 x^2-377/750 x^3+1/2500 x^4+\ldots

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maclaurin ln(1/(x+5))+xcos(x)

maclaurin

ln (\frac{1}{x + 5}) + x cos (x)

- ln (5) + \frac{4}{5} x + \frac{1}{50} x^{2} - \frac{377}{750} x^{3} + \frac{1}{2500} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= - ln (5) + \frac{\frac{d}{d x} ( ln ( \frac{1}{x + 5} ) + x cos ( x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( \frac{1}{x + 5} ) + x cos ( x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( \frac{1}{x + 5} ) + x cos ( x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= - ln (5) + \frac{\frac{4}{5}}{1 !} x + \frac{\frac{1}{25}}{2 !} x^{2} + \frac{- \frac{377}{125}}{3 !} x^{3} + \frac{\frac{6}{625}}{4 !} x^{4} + \dots

Refine

= - ln (5) + \frac{4}{5} x + \frac{1}{50} x^{2} - \frac{377}{750} x^{3} + \frac{1}{2500} x^{4} + \dots

\int sin (6 π t) d t

y = x - x^{3}, (1, 0)

\int 1 + tan (x) d x

\frac{d}{d x} (5 - 5 cos (x))

What is maclaurin ln(1/(x+5))+xcos(x) ?
The answer to maclaurin ln(1/(x+5))+xcos(x) is -ln(5)+4/5 x+1/50 x^2-377/750 x^3+1/2500 x^4+\ldots