Explain on step 2

You're asking about Step 2 of Part (ii), which is:

$\int \frac{1}{3 + 4y} dy = \int -\frac{1}{50} dt$

Here's the explanation for each side of the integral:

• Left side: $\int \frac{1}{3 + 4y} dy$ This is an integral of the form $\int \frac{1}{ax+b} dx$. The general solution for this type of integral is $\frac{1}{a} \ln|ax+b| + C$. In our case, $a=4$ and $b=3$. So, the integral becomes $\frac{1}{4} \ln|3 + 4y|$. (Alternatively, you can use a substitution: Let $u = 3+4y$, then $du = 4dy$, so $dy = \frac{1}{4}du$. The integral becomes $\int \frac{1}{u} \frac{1}{4} du = \frac{1}{4} \int \frac{1}{u} du = \frac{1}{4} \ln|u| = \frac{1}{4} \ln|3+4y|$.)

• Right side: $\int -\frac{1}{50} dt$ This is the integral of a constant with respect to $t$. The integral of a constant $c$ is $ct$. Here, the constant is $-\frac{1}{50}$. So, the integral becomes $-\frac{1}{50} t$.

Combining these results and adding a single constant of integration ($C$): $\frac{1}{4} \ln|3 + 4y| = -\frac{1}{50} t + C$ Then, to simplify, we multiply the entire equation by $4$: $\ln|3 + 4y| = -\frac{4}{50} t + 4C$ $\ln|3 + 4y| = -\frac{2}{25} t + C_1$ where $C_1$ is a new constant representing $4C$.