What is the integral of (arctan(7x))/(1+49x^2) ?

The integral of (arctan(7x))/(1+49x^2) is 1/14 arctan^2(7x)+C

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integral of (arctan(7x))/(1+49x^2)

\int \frac{arctan ( 7 x )}{1 + 49 x ^{2}} d x

\frac{1}{14} arctan^{2} (7 x) + C

Solution steps

\int \frac{arctan ( 7 x )}{1 + 49 x ^{2}} d x

Apply Integral Substitution

= \int \frac{arctan ( u )}{7 ( 1 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{arctan ( u )}{1 + u ^{2}} d u

Apply u-substitution

: \int v d v

= \frac{1}{7} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{7} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{7} \cdot \frac{arctan ^{2} ( 7 x )}{2}

Simplify

\frac{1}{7} \cdot \frac{arctan ^{2} ( 7 x )}{2} : \frac{1}{14} arctan^{2} (7 x)

= \frac{1}{14} arctan^{2} (7 x)

Add a constant to the solution

= \frac{1}{14} arctan^{2} (7 x) + C

x \to \frac{π}{3} lim (1 - 2 cos (x))

\frac{5}{x}, at x = 4

\frac{d y}{d x} = \frac{3 x + 2 y + 1}{3 x + 2 y - 1}

f (x) = (4 x^{2} - 8 x + 8) e^{x}

\frac{\partial}{\partial y} (x e^{2 y} sin (7 z))

What is the integral of (arctan(7x))/(1+49x^2) ?
The integral of (arctan(7x))/(1+49x^2) is 1/14 arctan^2(7x)+C