   Chapter 8.5, Problem 51E

Chapter
Section
Textbook Problem

# Using Two Methods Evaluate ∫ 0 1 x 1 + x 4 d x in two different ways, one of which is partial fractions.

To determine

To calculate: The integration, 01x1+x4dx in two different ways, one of which is partial fractions.

Explanation

Given:

The integral is 01x1+x4dx.

Formula used:

1. 1a2+u2du=1aarctanua+C

2. [f(u)±g(u)]du=f(u)du±g(u)du

Calculation:

Consider the following integral,

x1+x4dx

Substitute x2=u and square both side to get,

x4=u2

Differentiate x2=u with respect to u to get,

2xdx=duxdx=du2

Now substitute the above values we get,,

x1+x4dx=du21+u2=1211+u2du

Now use the basic rule of integration

1a2+u2du=1aarctanua+C

Hence,

x1+x4dx=12[11arctanu1]+C=12arctanu+C

Now replace the value of u to get,

x1+x4dx=12arctan(x2)+C

Now apply the limits and solve further,

01x1+x4dx=[12arctan(x2)]01=12[arctan(12)arctan(02)]=12π4

Thus,

01x1+x4dx=π8.

Second method: Solution by the method of partial fractions

Consider the following integral,

x1+x4dx

Since,

1+x4=(x2+2x+1)(x22x+1)

Now by using partial fraction method we get,

x1+x4=Ax+Bx2+2x+1+Cx+Dx22x+1

Simplify above equation by multiplying with the least common denominator, to get

x=(Ax+B)(x22x+1)+(Cx+D)(x2+2x+1)=Ax32Ax2+Ax+Bx22Bx+B+Cx3+2Cx2+Cx+Dx2+2Dx+D=(A+C)x3+(2A+B+2C+D)x2+(A2B+C+2D)x+(B+D)

Now, calculate value of A, B, C and D in above equation,

At x=0.

B+D=0B=D

Equate co-efficient of x3 on both sides.

A+C=0A=C

Equate co-efficient of x2.

2A+B+2C+D=02(C)+(D)+2C+D=022C=0.

C=0 and A=0

Equate co-efficient of x.

A2B+C+2D=102B+0+2D=122B=1

Hence,

B=122=122×22=24

and

D=122=122×22=24

So, by substituting the value of A, B, C and D we get,

x1+x4=(0)x24x2+2x+1+(0)x+24x22x+1=24(x2+2x+1)+24(x22x+1)

By solving further, we get,

x1+x4dx=(24(x2+2x+1)+24(x22x+1))dx=24(1(x2+2x+1)1(x22x+1))dx

Use the basic integration rule:

[f(u)±g(u)]du=f(u)du±g(u)du

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