What is the integral of 5arctan(4x) ?

The integral of 5arctan(4x) is 5(xarctan(4x)-1/8 ln|16x^2+1|)+C

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integral of 5arctan(4x)

\int 5 arctan (4 x) d x

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

Solution steps

\int 5 arctan (4 x) d x

Take the constant out:

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

= 5 \cdot \int arctan (4 x) d x

Apply Integration By Parts

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

= 5 (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

= 5 (x arctan (4 x) - \frac{1}{8} ln 16 x^{2} + 1)

Add a constant to the solution

= 5 (x arctan (4 x) - \frac{1}{8} ln 16 x^{2} + 1) + C

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What is the integral of 5arctan(4x) ?
The integral of 5arctan(4x) is 5(xarctan(4x)-1/8 ln|16x^2+1|)+C