Where does this $1/4$ come from?

Question

I am taking the improper integral of this function and I don't know where the $1/4$ comes form in this equation:

$$\int^0_{-\infty} \frac{1}{3-4x}dx$$

I know it turns into $\ln3-4x$ they also got $-\frac{1}{4}$ in front of the logarithm. How?

Answer 1

Hint: Try the change of variables $u=3-4x$ then $du=-4dx$ and....

Answer 2

Note the chain rule:

$$\frac d{dx}\ln|3-4x|=\frac{(3-4x)'}{3-4x}=\frac{-4}{3-4x}$$

To remove the $-4$, divide it from both sides:

$$\frac d{dx}\frac{\ln|3-4x|}{-4}=\frac1{3-4x}$$

Answer 3

sing $u$ substitution with $u = 3-4x$, $du = -4$, we get $$\int \frac{1}{3-4x} dx = -\frac{1}{4}\int\frac{-4 \,dx}{3-4x} = -\frac{1}{4}\int \frac{1}{u}du = -\frac{1}{4} \ln(u)+C$$ In general, you can just remember that if $\int f(t) \,dt = F(t)+C$, then $$\int f(at+b)\, dt = \frac{1}{a}F(at+b)+C$$