What is the integral of aarctan(bx) ?

The integral of aarctan(bx) is a(xarctan(bx)-1/(2b)ln|b^2x^2+1|)+C

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integral of aarctan(bx)

\int a arctan (b x) d x

a (x arctan (b x) - \frac{1}{2 b} ln b^{2} x^{2} + 1) + C

Solution steps

\int a arctan (b x) d x

Take the constant out:

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

= a \cdot \int arctan (b x) d x

Apply Integration By Parts

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

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

\int \frac{b x}{b ^{2} x ^{2} + 1} d x = \frac{1}{2 b} ln b^{2} x^{2} + 1

= a (x arctan (b x) - \frac{1}{2 b} ln b^{2} x^{2} + 1)

Add a constant to the solution

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

x \to 1 lim (x^{2} + 4)

y = - 0.02 x^{2} - 0.6 x + 90

\int 5 ln (x) d x

\frac{d}{d x} (\frac{1}{( x ^{2} + 1 ) ^{3}})

\int \frac{e ^{3 x}}{e ^{3 x} + 9} d x

What is the integral of aarctan(bx) ?
The integral of aarctan(bx) is a(xarctan(bx)-1/(2b)ln|b^2x^2+1|)+C