Calculus I Integral problem with Constant Numerator and Variable Denominator

Question

I have the following problem:

I believe I should use substitution. But I'm stuck. should u be set equal to the denominator? If so, what happens next?

It would be helpful to have to have a step by step breakdown

Answer 1

Use $(\tan^{-1}x)'=\frac1{1+x^2}$ to integrate as follows,

$$\int \frac6{7+y^2}dy=\frac6{\sqrt7}\int\frac{d(\frac y{\sqrt7})}{1+(\frac y{\sqrt7})^2}=\frac6{\sqrt7}\tan^{-1}(\frac y{\sqrt7})+C$$

Answer 2

think about this, the integral

$\int\frac{1}{1+x^2}dx$

the way to solve it, the substitution

x = tan($\theta$)

is the way to go. How can you get your integral to look somewhat like the integral above and what manipulations of the denominator would allow you to use that substitution

Answer 3

U-substitution won't be needed in the way that you expect.

This is actually a standard integral related to the function $\text{arctan}(x)$ (or $\text{tan}^{-1}(x)$), whose derivative can be shown to be $\frac{1}{1+x^2}$:

$$\theta = \text{arctan}(x)$$ $$\text{tan}(\theta) = x$$ $$\text{sec}^2(\theta) \frac{d\theta}{dx} = 1$$ $$\frac{d\theta}{dx} = \frac{d}{dx}\text{arctan}(x)= \frac{1}{\text{sec}^2(x)}=\frac{1}{1+\text{tan}^2(x)} = \frac{1}{1+x^2}$$

Using this knowledge, you can pull the 6 out of the integral, then use a u-substitution to pull out the 7 (i.e. $u=y/\sqrt{7}$), in order to find:

$$\int\frac{6}{7+y^2}dy= 6\int\frac{1}{7+7u^2} (\sqrt{7}du)=\frac{6}{\sqrt{7}}\int\frac{1}{1+u^2}du = \boxed{\frac{6}{\sqrt{7}}\text{arctan}(\frac{y}{\sqrt{7}}) + C}$$