What is the integral of 1/x+4cos(5/x) ?

The integral of 1/x+4cos(5/x) is ln|x|-20(-1/5 xcos(5/x)-\Si(5/x))+C

Popular >

integral of 1/x+4cos(5/x)

\int \frac{1}{x} + 4 cos (\frac{5}{x}) d x

ln ∣ x ∣ - 20 (- \frac{1}{5} x cos (\frac{5}{x}) - Si (\frac{5}{x})) + C

Solution steps

\int \frac{1}{x} + 4 cos (\frac{5}{x}) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x} d x + \int 4 cos (\frac{5}{x}) d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int 4 cos (\frac{5}{x}) d x = - 20 (- \frac{1}{5} x cos (\frac{5}{x}) - Si (\frac{5}{x}))

= ln ∣ x ∣ - 20 (- \frac{1}{5} x cos (\frac{5}{x}) - Si (\frac{5}{x}))

Add a constant to the solution

= ln ∣ x ∣ - 20 (- \frac{1}{5} x cos (\frac{5}{x}) - Si (\frac{5}{x})) + C

\int \frac{x}{( cos ( x ^{2} ) ) ^{2}} d x

\int - 5 x^{2} (3 x - 8)^{4} d x

\int \frac{2}{2 - x ^{2}} d x

\int \frac{1}{1 + \frac{1}{x ^{2}}} d x

\int \frac{sec ^{2} ( 5 x )}{1 - tan ( 5 x )} d x

What is the integral of 1/x+4cos(5/x) ?
The integral of 1/x+4cos(5/x) is ln|x|-20(-1/5 xcos(5/x)-\Si(5/x))+C