Integral of arctg((3x)/4) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \frac{3 x}{4}$.

      Then let $du = \frac{3 dx}{4}$ and substitute $\frac{4 du}{3}$:

      $$\int \frac{16 \operatorname{atan}{\left(u \right)}}{9}\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{4 \operatorname{atan}{\left(u \right)}}{3}\, du = \frac{4 \int \operatorname{atan}{\left(u \right)}\, du}{3}$$

         1. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let $u{\left(u \right)} = \operatorname{atan}{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = 1$.

            Then $\operatorname{du}{\left(u \right)} = \frac{1}{u^{2} + 1}$.

            To find $v{\left(u \right)}$:

            1. The integral of a constant is the constant times the variable of integration:

               $$\int 1\, du = u$$

            Now evaluate the sub-integral.

         2. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \frac{u}{u^{2} + 1}\, du = \frac{\int \frac{2 u}{u^{2} + 1}\, du}{2}$$

            1. Let $u = u^{2} + 1$.

               Then let $du = 2 u du$ and substitute $\frac{du}{2}$:

               $$\int \frac{1}{2 u}\, du$$

               1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

               Now substitute $u$ back in:

               $$\log{\left(u^{2} + 1 \right)}$$

            So, the result is: $\frac{\log{\left(u^{2} + 1 \right)}}{2}$

         So, the result is: $\frac{4 u \operatorname{atan}{\left(u \right)}}{3} - \frac{2 \log{\left(u^{2} + 1 \right)}}{3}$

      Now substitute $u$ back in:

      $$x \operatorname{atan}{\left(\frac{3 x}{4} \right)} - \frac{2 \log{\left(\frac{9 x^{2}}{16} + 1 \right)}}{3}$$

   Method #2

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(x \right)} = \operatorname{atan}{\left(\frac{3 x}{4} \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

      Then $\operatorname{du}{\left(x \right)} = \frac{3}{4 \cdot \left(\frac{9 x^{2}}{16} + 1\right)}$.

      To find $v{\left(x \right)}$:

      1. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, dx = x$$

      Now evaluate the sub-integral.

   2. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \frac{3 x}{4 \cdot \left(\frac{9 x^{2}}{16} + 1\right)}\, dx = \frac{3 \int \frac{x}{\frac{9 x^{2}}{16} + 1}\, dx}{4}$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{x}{\frac{9 x^{2}}{16} + 1}\, dx = \frac{8 \int \frac{9 x}{8 \cdot \left(\frac{9 x^{2}}{16} + 1\right)}\, dx}{9}$$

         1. Let $u = \frac{9 x^{2}}{16} + 1$.

            Then let $du = \frac{9 x dx}{8}$ and substitute $\frac{8 du}{9}$:

            $$\int \frac{8}{9 u}\, du$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            Now substitute $u$ back in:

            $$\log{\left(\frac{9 x^{2}}{16} + 1 \right)}$$

         So, the result is: $\frac{8 \log{\left(\frac{9 x^{2}}{16} + 1 \right)}}{9}$

      So, the result is: $\frac{2 \log{\left(\frac{9 x^{2}}{16} + 1 \right)}}{3}$

2. Now simplify:

   $$x \operatorname{atan}{\left(\frac{3 x}{4} \right)} - \frac{2 \log{\left(\frac{9 x^{2}}{16} + 1 \right)}}{3}$$

3. Add the constant of integration:

   $$x \operatorname{atan}{\left(\frac{3 x}{4} \right)} - \frac{2 \log{\left(\frac{9 x^{2}}{16} + 1 \right)}}{3}+ \mathrm{constant}$$

The answer is:

$$x \operatorname{atan}{\left(\frac{3 x}{4} \right)} - \frac{2 \log{\left(\frac{9 x^{2}}{16} + 1 \right)}}{3}+ \mathrm{constant}$$