What is the integral of (41arctan(x))/(x^2) ?

The integral of (41arctan(x))/(x^2) is 41(-(arctan(x))/x+ln|x|-1/2 ln|x^2+1|)+C

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integral of (41arctan(x))/(x^2)

\int \frac{41 arctan ( x )}{x ^{2}} d x

41 (- \frac{arctan ( x )}{x} + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) + C

Solution steps

\int \frac{41 arctan ( x )}{x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 41 \cdot \int \frac{arctan ( x )}{x ^{2}} d x

Apply Integration By Parts

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

= 41 (- \frac{arctan ( x )}{x} - \int - \frac{1}{x ( x ^{2} + 1 )} d x)

\int - \frac{1}{x ( x ^{2} + 1 )} d x = - ln ∣ x ∣ + \frac{1}{2} ln x^{2} + 1

= 41 (- \frac{arctan ( x )}{x} - (- ln ∣ x ∣ + \frac{1}{2} ln x^{2} + 1))

Simplify

= 41 (- \frac{arctan ( x )}{x} + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1)

Add a constant to the solution

= 41 (- \frac{arctan ( x )}{x} + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) + C

f (x) = sin (ln (x))

\frac{1}{x + 2}

\frac{d}{d x} (cos (e^{- 5 x^{2}}))

x \to 0 lim ((1 - 5 x)^{\frac{5}{x}})

\int cos (a x) d x

What is the integral of (41arctan(x))/(x^2) ?
The integral of (41arctan(x))/(x^2) is 41(-(arctan(x))/x+ln|x|-1/2 ln|x^2+1|)+C