Integration of some special irrational algebraic functions

In this section, we shall discuss four integrals of the form $\displaystyle \int { \frac { \phi \left( x \right) }{ P\sqrt { Q } } } dx$ when P and Q are polynomial functions of x.

Type I

Integrals of the form $\displaystyle \int { \frac { \phi \left( x \right) }{ P\sqrt { Q } } } dx$ where P and Q both are linear function of x

To evaluate this type of integrals we put $\displaystyle Q={ t }^{ 2 }$ i.e., to evaluate integrals of the form $\displaystyle \int { \frac { 1 }{ \left( ax+b \right) \sqrt { cx+d } } } dx\quad put\quad cx+d={ t }^{ 2 }$

The following examples illustrate the procedure.

Example

$\displaystyle \int { \frac { 1 }{ \left( x-3 \right) \sqrt { x+1 } } } dx$

Solution:

$\displaystyle Let\quad I=\int { \frac { 1 }{ \left( x-3 \right) \sqrt { x+1 } } } dx$

Here P and Q both are linear, so we put

$\displaystyle Q={ t }^{ 2 }\quad i.e.,\quad x+1={ t }^{ 2 }\quad so\quad that\quad dx=2t\quad dt\\ \therefore I=\int { \frac { 1 }{ \left( { t }^{ 2 }-1-3 \right) } \frac { 2t }{ \sqrt { { t }^{ 2 } } } } dt\\ =2\int { \frac { dt }{ { t }^{ 2 }-{ 2 }^{ 2 } } } \\ =2.\frac { 1 }{ 2\left( 2 \right) } \log { \left| \frac { t-2 }{ t+2 } \right| } +C\\ =\frac { 1 }{ 2 } \log { \left| \frac { \sqrt { x+1 } -2 }{ \sqrt { x+1 } +2 } \right| } +C$

Type II

Integrals of the form $\displaystyle \int { \frac { \phi \left( x \right) }{ P\sqrt { Q } } } dx$ where P is a quadratic expression and Q is a linear expression

To evaluate this type of integrals we put $\displaystyle Q={ t }^{ 2 }$ i.e. to evaluate integrals of the form $\displaystyle \int { \frac { 1 }{ \left( a{ x }^{ 2 }+bx+c \right) \sqrt { px+q } } } dx,\quad put\quad px+q={ t }^{ 2 }$

Example

$\displaystyle \int { \frac { 1 }{ \left( { x }^{ 2 }-4 \right) \sqrt { x+1 } } } dx$

Solution:

$\displaystyle Let\quad I=\int { \frac { 1 }{ \left( { x }^{ 2 }-4 \right) \sqrt { x+1 } } } dx\\ Put\quad x+1={ t }^{ 2 }\quad so\quad that\quad dx=2t\quad dt$

$\displaystyle \therefore I=\int { \frac { 2t }{ \left[ { \left( { t }^{ 2 }-1 \right) }^{ 2 }-4 \right] \sqrt { { t }^{ 2 } } } dt } \\ =2\int { \frac { dt }{ \left( { t }^{ 2 }-1-2 \right) \left( { t }^{ 2 }-1+2 \right) } } \\ =2\int { \frac { dt }{ \left( { t }^{ 2 }-3 \right) \left( { t }^{ 2 }+1 \right) } }$

$\displaystyle Let\quad { t }^{ 2 }=y\quad then\\ \frac { 1 }{ \left( { t }^{ 2 }-3 \right) \left( { t }^{ 2 }+1 \right) } =\frac { 1 }{ \left( y-3 \right) \left( y+1 \right) } \\ \frac { 1 }{ \left( y-3 \right) \left( y+1 \right) } =\frac { 1 }{ \left( y-3 \right) } +\frac { B }{ \left( y+1 \right) } \qquad .......\left( i \right) \\ \Rightarrow 1=A\left( y+1 \right) +B\left( y-3 \right) \qquad .......\left( ii \right)$

Putting y=-1 , 3 respectively in (ii) , we get $\displaystyle B=-\frac { 1 }{ 4 } \quad and\quad A=\frac { 1 }{ 4 }$

Substituting the values of A and B in (i) we obtain

$\displaystyle \frac { 1 }{ \left( y-3 \right) \left( y+1 \right) } =\frac { 1 }{ 4\left( y-3 \right) } -\frac { 1 }{ 4\left( y+1 \right) } \\ \Rightarrow \frac { 1 }{ \left( { t }^{ 2 }-3 \right) \left( { t }^{ 2 }+1 \right) } =\frac { 1 }{ 4\left( { t }^{ 2 }-3 \right) } -\frac { 1 }{ 4\left( { t }^{ 2 }+1 \right) }$

$\displaystyle So,\quad I=2\int { \left( \frac { 1 }{ 4\left( { t }^{ 2 }-3 \right) } -\frac { 1 }{ 4\left( { t }^{ 2 }+1 \right) } \right) } dt\\ =\frac { 1 }{ 2 } \int { \frac { 1 }{ { t }^{ 2 }-{ \left( \sqrt { 3 } \right) }^{ 2 } } dt } -\frac { 1 }{ 2 } \int { \frac { 1 }{ { t }^{ 2 }+{ 1 }^{ 2 } } } dt$

$\displaystyle =\frac { 1 }{ 2 } .\frac { 1 }{ 2\sqrt { 3 } } \log { \left| \frac { t-\sqrt { 3 } }{ t+\sqrt { 3 } } \right| } -\frac { 1 }{ 2 } \tan ^{ -1 }{ \left( t \right) } +C\\ =\frac { 1 }{ 4\sqrt { 3 } } \log { \left| \frac { \sqrt { x+1 } -\sqrt { 3 } }{ \sqrt { x+1 } +\sqrt { 3 } } \right| } -\frac { 1 }{ 2 } \tan ^{ -1 }{ \left( \frac { \sqrt { x+1 } }{ 1 } \right) } +C$

Type III

Integrals of the form $\displaystyle \int { \frac { \phi \left( x \right) }{ P\sqrt { Q } } } dx$ where P is a linear expression and Q is a quadratic expression

To evaluate this type of integrals we put $\displaystyle p=\frac { 1 }{ t }$ i.e. to evaluate integrals of the form $\displaystyle \int { \frac { 1 }{ \left( ax+b \right) \sqrt { p{ x }^{ 2 }+qx+r } } } dx\quad put\quad ax+b=\frac { 1 }{ t }$

Example

$\displaystyle \int { \frac { 1 }{ \left( x+1 \right) \sqrt { { x }^{ 2 }-1 } } } dx$

Solution:

$\displaystyle Let\quad I=\int { \frac { 1 }{ \left( x+1 \right) \sqrt { { x }^{ 2 }-1 } } } dx\quad Put\quad x+1=\frac { 1 }{ t } \quad so\quad that\quad \\dx=-\frac { 1 }{ { t }^{ 2 } } dt$

$\displaystyle \therefore I=\int { \frac { 1 }{ \frac { 1 }{ t } \sqrt { { \left( \frac { 1 }{ t } -1 \right) }^{ 2 }-1 } } } .\left( -\frac { 1 }{ { t }^{ 2 } } \right) dt\\ =-\int { \frac { dt }{ \sqrt { 1-2t } } } \\ =-\int { { \left( 1-2t \right) }^{ -\frac { 1 }{ 2 } } } dt$

$\displaystyle =-\frac { { \left( 1-2t \right) }^{ \frac { 1 }{ 2 } } }{ \left( -2 \right) \left( \frac { 1 }{ 2 } \right) } +C\\ =\sqrt { 1-2t } +C\\ =\sqrt { 1-\frac { 2 }{ x+1 } } +C\\ =\sqrt { \frac { x-1 }{ x+1 } } +C$

Type IV

Integrals of the form $\displaystyle \int { \frac { \phi \left( x \right) }{ P\sqrt { Q } } } dx$ where P and Q both are pure quadratic expression in x i.e. $\displaystyle P=a{ x }^{ 2 }+b\quad and\quad Q=c{ x }^{ 2 }+d$

To evaluate this type of integrals we put $\displaystyle x=\frac { 1 }{ t }$ and then $\displaystyle c+d{ t }^{ 2 }={ u }^{ 2 }$ i.e., to evaluate integrals of the form

$\displaystyle \int { \frac { 1 }{ \left( a{ x }^{ 2 }+b \right) \sqrt { c{ x }^{ 2 }+d } } } dx\quad we\quad put\quad x=\frac { 1 }{ t } \quad to\quad obtain\quad \int { \frac { -tdt }{ \left( a+b{ t }^{ 2 } \right) \sqrt { c+d{ t }^{ 2 } } } } \quad and\quad then\quad \\ c+d{ t }^{ 2 }={ u }^{ 2 }$

Example

$\displaystyle \int { \frac { 1 }{ { x }^{ 2 }\sqrt { 1+{ x }^{ 2 } } } } dx$

Solution:

$\displaystyle Let\quad I=\int { \frac { 1 }{ { x }^{ 2 }\sqrt { 1+{ x }^{ 2 } } } } dx\quad Put\quad x=\frac { 1 }{ t } \quad so\quad that\quad -\frac { 1 }{ { x }^{ 2 } } dx=dt\quad or\quad dx=-{ x }^{ 2 }dt$

$\displaystyle \therefore I=\int { \frac { -dt }{ \sqrt { 1+\frac { 1 }{ { t }^{ 2 } } } } } \\ =-\int { \frac { t\quad dt }{ \sqrt { { t }^{ 2 }+1 } } } \\ =-\int { \frac { u\quad du }{ \sqrt { { u }^{ 2 } } } } \\ where\quad { t }^{ 2 }+1={ u }^{ 2 }$

$\displaystyle =\int { -1 } du\\ =-u+C\\ =-\sqrt { { t }^{ 2 }+1 } +C\\ =-\sqrt { \frac { 1 }{ { x }^{ 2 } } +1 } +C\\ =-\frac { \sqrt { 1+{ x }^{ 2 } } }{ x } +C$

April 18, 2019

Integration

Integration of some special irrational algebraic functions

Type I

Example

Type II

Example

Type III

Example

Type IV

Example

Limits

Continuity

Differentiability

Differentiation

Tangents and Normals

Increasing and Decreasing Functions

Maxima and Minima

Indefinite Integrals

Definite Integrals

Improper integral

Differential Equations

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