Ucale

# 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
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