Q. 1 - Calculus

Question

Solve each of the following indefinite integrals 

a)$\int \frac{1}{y^2}dy$

b)$\int \frac{1}{1-y}dy$

c)$\int \frac{1}{y^2+1}dy$

d)$\int \frac{y}{y^2+1}dy$

e)$\int \frac{1}{y(1-y)}dy$

f)$\int \frac{1}{y(1-3y)}dy$

Step-by-Step Solution

Answer

The integral is:

$$\begin{gathered} a)-\frac{1}{y}+c \\ b)-\ln (1-y)+c \\ c){\tan }^{-1}y+c \\ d)\frac{1}{2}\ln (y^2+1)+c \\ e)\ln \left(\frac{y}{1-y}\right)+c \\ f)\ln \left(\frac{y}{1-3y}\right)+c \end{gathered}$$

Part (a) Step 1. Given information

The integrals to be solved are:

$$\begin{gathered} \int \frac{1}{y^2}dy, \\ \int \frac{1}{1-y}dy, \\ \int \frac{1}{y^2+1}dy \\ \int \frac{y}{y^2+1}dy \\ \int \frac{1}{y(1-y)}dy \\ \int \frac{1}{y(1-3y)}dy \end{gathered}$$

Part (a) step 2. Solve integral:

$$\begin{gathered} \int \frac{1}{y^2}dy \\ =\int y^{-2}dy \\ =\frac{y^{-2+1}}{-2+1}+c \\ =\frac{y^{-1}}{-1}+c \\ =-\frac{1}{y}+c \end{gathered}$$

Where c is a constant.

Part (b) step 1. Solve integral:

$\int \frac{1}{1-y}dy$

Let $$\begin{gathered} z=1-y \\ dz=-dy \end{gathered}$$

so,$$\begin{gathered} \int \frac{1}{1-y}dy \\ =\int -\frac{1}{z}dz \\ =-\ln z+c \\ =\ln (1-y)+c \end{gathered}$$

Part (c) step 1. Solve integral:

$$\begin{gathered} \int \frac{1}{1+y^2}dy \\ ={\tan }^{-1}y+c \end{gathered}$$

Part (d) step 1. Solve integral:

$\int \frac{y}{y^2+1}dy$

let $$\begin{gathered} z=y^2+1 \\ dz=2ydy \\ ydy=\frac{1}{2}dz \end{gathered}$$

So,

$$\begin{gathered} \int \frac{y}{y^2+1}dy \\ =\int \frac{dz}{2z} \\ =\frac{1}{2}\ln z+c \\ =\frac{1}{2}\ln (1+y^2)+c \end{gathered}$$

Part (e) step 1. Solve integral:

$$\begin{gathered} \int \frac{1}{y(1-y)}dy \\ =\int \left(\frac{1}{y}+\frac{1}{1-y}\right)dy \\ =\int \frac{1}{y}dy+\int \frac{1}{1-y}dy \\ =\ln y-\ln (1-y)+c \\ =\ln \left(\frac{y}{1-y}\right)+c \end{gathered}$$

Part (f) step 1. Solve integral:

$$\begin{gathered} \int \frac{1}{y(1-3y)}dy \\ =\int \left(\frac{1}{y}+\frac{3}{1-3y}\right)dy \\ =\int \frac{1}{y}dy+\int \frac{3}{1-3y}dy \\ =\ln y+\int \frac{3}{1-3y}dy \end{gathered}$$

Let  $$\begin{gathered} 1-3y=z \\ -3dy=dz \\ 3dy=-dz \end{gathered}$$

So, $$\begin{gathered} \ln y+\int \frac{3}{1-3y}dy \\ =\ln y-\int \frac{dz}{z} \\ =\ln y-\ln z+c \\ =\ln y-\ln (1-3y)+c \\ =\ln \left(\frac{y}{1-3y}\right)+c \end{gathered}$$