Evaluate the Integral

Problem

(∫_1^2)(1/(1+x4)*d(x))

Solution

1. Factor the denominator using the Sophie Germain identity to prepare for partial fraction decomposition.

1+x4=(x2+1)2−2*x2

1+x4=(x2+√(,2)*x+1)*(x2−√(,2)*x+1)

2. Decompose the integrand into partial fractions.

1/(1+x4)=(A*x+B)/(x2+√(,2)*x+1)+(C*x+D)/(x2−√(,2)*x+1)

1/(1+x4)=1/(2√(,2))*((x+√(,2))/(x2+√(,2)*x+1)−(x−√(,2))/(x2−√(,2)*x+1))

3. Rewrite the numerators to facilitate integration by creating terms that match the derivative of the denominators.

(x+√(,2))/(x2+√(,2)*x+1)=1/2(2*x+√(,2))/(x2+√(,2)*x+1)+√(,2)/21/((x+√(,2)/2)2+1/2)

(x−√(,2))/(x2−√(,2)*x+1)=1/2(2*x−√(,2))/(x2−√(,2)*x+1)−√(,2)/21/((x−√(,2)/2)2+1/2)

4. Integrate each term using the natural logarithm and arctangent rules.

(∫_^)(1/(1+x4)*d(x))=1/(4√(,2))*ln((x2+√(,2)*x+1)/(x2−√(,2)*x+1))+1/(2√(,2))*(arctan(√(,2)*x+1)+arctan(√(,2)*x−1))

5. Apply the fundamental theorem of calculus by evaluating the antiderivative at the limits x=2 and x=1

[1/(4√(,2))*ln((x2+√(,2)*x+1)/(x2−√(,2)*x+1))+1/(2√(,2))*arctan((√(,2)*x)/(1−x2))]21

6. Simplify the result by plugging in the values. Note that for the arctangent term, we account for the branch cut as the denominator 1−x2 passes through zero.

Value at *2=1/(4√(,2))*ln((5+2√(,2))/(5−2√(,2)))+1/(2√(,2))*arctan(−(2√(,2))/3)

Value at *1=1/(4√(,2))*ln((2+√(,2))/(2−√(,2)))+1/(2√(,2))π/2

Final Answer

(∫_1^2)(1/(1+x4)*d(x))=1/(4√(,2))*ln(((5+2√(,2))*(2−√(,2)))/((5−2√(,2))*(2+√(,2))))+1/(2√(,2))*(π+arctan(−(2√(,2))/3)−π/2)

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