Integral of 3/sqrt(6x-5)+7/x^2 dx

1. Integrate term-by-term:

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

      $$\int \frac{3}{\sqrt{6 x - 5}}\, dx = 3 \int \frac{1}{\sqrt{6 x - 5}}\, dx$$

      1. Let $u = \sqrt{6 x - 5}$.

         Then let $du = \frac{3 dx}{\sqrt{6 x - 5}}$ and substitute $\frac{du}{3}$:

         $$\int \frac{1}{3}\, du$$

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

            $$\text{False}$$

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

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

            So, the result is: $\frac{u}{3}$

         Now substitute $u$ back in:

         $$\frac{\sqrt{6 x - 5}}{3}$$

      So, the result is: $\sqrt{6 x - 5}$

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

      $$\int \frac{7}{x^{2}}\, dx = 7 \int \frac{1}{x^{2}}\, dx$$

      PiecewiseRule(subfunctions=[(ArctanRule(a=1, b=1, c=0, context=1/(x**2), symbol=x), False), (ArccothRule(a=1, b=1, c=0, context=1/(x**2), symbol=x), False), (ArctanhRule(a=1, b=1, c=0, context=1/(x**2), symbol=x), False)], context=1/(x**2), symbol=x)

      So, the result is: $\text{NaN}$

   The result is: $\text{NaN}$

2. Add the constant of integration:

   $$\text{NaN}+ \mathrm{constant}$$

The answer is:

$$\text{NaN}+ \mathrm{constant}$$