What is the integral of arctan(4x) ?

The integral of arctan(4x) is xarctan(4x)-1/8 ln|16x^2+1|+C

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integral of arctan(4x)

\int arctan (4 x) d x

x arctan (4 x) - \frac{1}{8} ln 16 x^{2} + 1 + C

Solution steps

\int arctan (4 x) d x

Apply Integration By Parts

: x arctan (4 x) - \int \frac{4 x}{16 x ^{2} + 1} d x

= x arctan (4 x) - \int \frac{4 x}{16 x ^{2} + 1} d x

\int \frac{4 x}{16 x ^{2} + 1} d x = \frac{1}{8} ln 16 x^{2} + 1

= x arctan (4 x) - \frac{1}{8} ln 16 x^{2} + 1

Add a constant to the solution

= x arctan (4 x) - \frac{1}{8} ln 16 x^{2} + 1 + C

f (x) = \frac{1 - x}{2 + x}

x \to 2 lim (ln (x - 1))

\int cos^{7} (θ) sin (θ) d θ

\frac{1}{( 1 + x ) ^{3}}

ln (\frac{4 x + 3}{3 x - 5})

What is the integral of arctan(4x) ?
The integral of arctan(4x) is xarctan(4x)-1/8 ln|16x^2+1|+C