$$\int{\frac{3}{5y^2 + 4}}dy$$

$$\frac{3}{4}\int{\frac{1}{\left(\frac{\sqrt{5}y}{2}\right)^2 + 1}}dy$$

$$u = \frac{\sqrt{5}y}{2}$$

$$dy = \frac{2}{\sqrt{5}}du$$

My solution to this problem was

$$\frac{3}{2\sqrt{5}}\left(\frac{1}{\tan\left(\frac{\sqrt{5}y}{2}\right)} + c\right)$$

However, apparently the solution is as follows: $$\frac{3}{2\sqrt{5}}\left(\frac{1}{\tan\frac{\sqrt{5}y}{2}}\right) + c$$

My question is: why is c outside the bracket? I was taught that when you integrate a value you must add c. But since we're adding c, we also must multiply by whatever is outside of the integral.

-
It is just the same. Just expand your expression and you have another constant. –  Claude Leibovici Mar 14 at 6:28
Did anybody notice that the integral is wrong? –  alex Mar 14 at 7:11

Remember that $c$ is a generic constant - it doesn't hold a particular value until it is assigned one. In this case, note that $\frac{3}{2\sqrt{5}}*c$ is still a constant. $$\frac{3}{2\sqrt{5}}\left(\frac{1}{\tan(\frac{\sqrt{5}y}{2})} + c\right)=\frac{3}{2\sqrt{5}}\left(\frac{1}{\tan(\frac{\sqrt{5}y}{2})}\right)+\frac{3}{2\sqrt{5}}*c$$ $$\frac{3}{2\sqrt{5}}\left(\frac{1}{\tan(\frac{\sqrt{5}y}{2})}\right)+c$$

-

Doesn't matter if c is inside or outside the bracket. It just denotes a constant value. If you multiply the coefficient with the constant, you will just get another constant.

-

This is because what ever the value is that is being multiplied is still a constant. The mathematician does not concern themselves with any magnitude of this constant when solving empirically.

-