Chapter 6: Q. 1 (page 569)

Solve each of the following indefinite integrals
a) $\int \frac{1}{y^{2}} d y$
b) $\int \frac{1}{1 - y} d y$
c) $\int \frac{1}{y^{2} + 1} d y$
d) $\int \frac{y}{y^{2} + 1} d y$
e) $\int \frac{1}{y (1 - y)} d y$
f) $\int \frac{1}{y (1 - 3 y)} d y$

Short Answer

Expert verified

The integral is:

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

Step by step solution

Part (a) Step 1. Given information

The integrals to be solved are:

$\int \frac{1}{y^{2}} d y, \int \frac{1}{1 - y} d y, \int \frac{1}{y^{2} + 1} d y \int \frac{y}{y^{2} + 1} d y \int \frac{1}{y (1 - y)} d y \int \frac{1}{y (1 - 3 y)} d y$

Part (a) step 2. Solve integral:

$\int \frac{1}{y^{2}} d y = \int y^{- 2} d y = \frac{y^{- 2 + 1}}{- 2 + 1} + c = \frac{y^{- 1}}{- 1} + c = - \frac{1}{y} + c$

Where c is a constant.

Part (b) step 1. Solve integral:

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

Let $z = 1 - y d z = - d y$

so, $\int \frac{1}{1 - y} d y = \int - \frac{1}{z} d z = - \ln z + c = \ln (1 - y) + c$

Part (c) step 1. Solve integral:

$\int \frac{1}{1 + y^{2}} d y = \tan^{- 1} y + c$

Part (d) step 1. Solve integral:

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

let $z = y^{2} + 1 d z = 2 y d y y d y = \frac{1}{2} d z$

So,

$\int \frac{y}{y^{2} + 1} d y = \int \frac{d z}{2 z} = \frac{1}{2} \ln z + c = \frac{1}{2} \ln (1 + y^{2}) + c$

Part (e) step 1. Solve integral:

$\int \frac{1}{y (1 - y)} d y = \int (\frac{1}{y} + \frac{1}{1 - y}) d y = \int \frac{1}{y} d y + \int \frac{1}{1 - y} d y = \ln y - \ln (1 - y) + c = \ln (\frac{y}{1 - y}) + c$

Part (f) step 1. Solve integral:

$\int \frac{1}{y (1 - 3 y)} d y = \int (\frac{1}{y} + \frac{3}{1 - 3 y}) d y = \int \frac{1}{y} d y + \int \frac{3}{1 - 3 y} d y = \ln y + \int \frac{3}{1 - 3 y} d y$

Let $1 - 3 y = z - 3 d y = d z 3 d y = - d z$

So, $\ln y + \int \frac{3}{1 - 3 y} d y = \ln y - \int \frac{d z}{z} = \ln y - \ln z + c = \ln y - \ln (1 - 3 y) + c = \ln (\frac{y}{1 - 3 y}) + c$

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Solve each of the following indefinite integrals
a) $\int \frac{1}{y^{2}} d y$
b) $\int \frac{1}{1 - y} d y$
c) $\int \frac{1}{y^{2} + 1} d y$
d) $\int \frac{y}{y^{2} + 1} d y$
e) $\int \frac{1}{y (1 - y)} d y$
f) $\int \frac{1}{y (1 - 3 y)} d y$

Short Answer

Step by step solution

Part (a) Step 1. Given information

Part (a) step 2. Solve integral:

Part (b) step 1. Solve integral:

Part (c) step 1. Solve integral:

Part (d) step 1. Solve integral:

Part (e) step 1. Solve integral:

Part (f) step 1. Solve integral:

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Solve each of the following indefinite integralsa)∫1y2dyb)∫11-ydyc)∫1y2+1dyd)∫yy2+1dye)∫1y(1-y)dyf)∫1y(1-3y)dy

Short Answer

Step by step solution

Part (a) Step 1. Given information

Part (a) step 2. Solve integral:

Part (b) step 1. Solve integral:

Part (c) step 1. Solve integral:

Part (d) step 1. Solve integral:

Part (e) step 1. Solve integral:

Part (f) step 1. Solve integral:

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Most popular questions from this chapter

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Calculus

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Probability and Statistics

Study anywhere. Anytime. Across all devices.

Solve each of the following indefinite integrals
a) $\int \frac{1}{y^{2}} d y$
b) $\int \frac{1}{1 - y} d y$
c) $\int \frac{1}{y^{2} + 1} d y$
d) $\int \frac{y}{y^{2} + 1} d y$
e) $\int \frac{1}{y (1 - y)} d y$
f) $\int \frac{1}{y (1 - 3 y)} d y$