A calc 2 question on integration

Question

I have the integral $$\int \frac 5{90-2t} dt$$. I thought that's $5ln(90-2t)$. I thought as long as the $t$ is first degree the integral is $ln(denominator)$. Could someone just remind of the rule for this integral? I know I could use a u-substitution method but I want to know the clear rule for integrating when the variable is on the bottom.

Answer 1

The rule is

$$\int \dfrac {du}{u} = \ln |u| + C$$

If you let $u = 90-2t$ on the bottom, then $du = -2 dt$, which is not on the top, so you cannot apply to rule. You must manipulate the top to be $-2 dt$.

To do that, multiply the integral by a convenient form of $1$:

$$\int \frac 5{90-2t} dt= \dfrac {5}{-2}\cdot \dfrac {-2}{5} \int \frac 5{90-2t} dt = \dfrac {5}{-2} \int \frac {-2dt}{90-2t} =\dfrac {5}{-2} \ln |90-2t|+C$$

Answer 2

Remember the chain rule! 5$\int \frac{1}{u} du = \frac{5}{u’}ln(|u|)$

Answer 3

Here is the method:

$\int\frac{f'(x)}{f(x)}dx=ln(f(x))+C$ where $C$ is integral constant.

According to your problem:

$\int\frac{5}{90-2t}dt=\int-\frac{5}{2}*\frac{-2}{90-2t}dt=-\frac{5}{2}ln(90-2t)+C$, $C$ is the constant of integration.

Answer 4

$$\int \frac 5{90-2t} dt$$

$$ u=90-2t$$ $$du = -2dt$$ $$\int \frac 5{90-2t} dt = \int \frac {-5}{2} \frac {-2}{90-2t} dt= \frac {-5}{2} \int \frac {du}{u}= $$

$$\frac {-5}{2} ln|u|+C=$$

$$\frac {-5}{2} ln|90-2t|+C=$$