Evaluate the integral (step by step process)

Question

∫x⁴+X²+4/ x⁶+4x⁴+4x² dx

Answer 1

First do the partial fraction expansion:

(x⁴+x²+4)/(x⁶+4x⁴+4x²) =(x⁴+x²+4)/(x²(x²+2)²) = A/x + B/x² + (Cx+D)/(x²+2) + (Ex+F)/(x²+2)²

x⁴+x²+4 = Ax⁵+4Ax³+4Ax + Bx⁴+4Bx²+4B + Cx⁵+ 2Cx³+Dx⁴+ 2Dx²+Ex³+Fx²

x⁵: A+C=0

x⁴: B+D=1

x³: 4A+2C+E=0

x²: 4B+2C+2D+F=1

x: 4A=0

4: 4B = 4

A=0

B=1

C=0

D=0

E=0

F=-3

Which gives us two integrals

∫1/x²dx - 3 ∫1/(x²+2)² dx

The first is equal to -1/x

For the second, let u = √2tan(u), u = tan^-1(x/√2), dx = √2sec²(u) du

1/(4(tan²+1)²) = 1/(4sec⁴(u))

∫√2sec²(u)/(4sec⁴(u) du = (√2/4)∫1/sec²(u) du = (√2/4)∫cos²(u) du = (√2/4)∫(1+cos(2u))/2 du

(√2/8)(u+sin(2u)/2) = (√2/16)(2tan^-1(x/√2) + sin(2tan^-1(x/√2)))

Putting it all together:

∫(x⁴+x²+4)/(x⁶+4x⁴+4x²) dx = -1/x - 3(√2/16)(2tan^-1(x/√2) + sin(2tan^-1(x/√2))) + C

Answer 2

This is a fairly involved partial fractions problem. Hopefully these steps are clear:

First factor denominator:

(x⁴+x²+4)/(x²(x²+2)²)

Next we'll break this up as a partial fraction:

(x⁴+x²+4)/(x²(x²+2)²) = (Ax+B)/x² + (Cx+D)/(x²+2) + (Ex+F)/(x²+2)²

Multiply each side by common denominator to eliminate fractions which results in:

x⁴+x²+4 = (Ax+B)(x²+2)² + (Cx+D)x²(x²+2) + (Ex+F)x²

Now comes the fun part...multiply out the right side:

x⁴+x²+4 = Ax⁵+4Ax³+4Ax+Bx⁴+4Bx²+4B+Cx⁵+2Cx³+Dx⁴+2Dx²+Ex³+Fx²

Combine like terms and factor out the powers of x:

x⁴+x²+4 = (A+C)x⁵ + (B+D)x⁴ + (4A+2C+E)x³ + (4B+2D+F)x² + 4Ax + 4B

Equating terms on the left and right side results in the following system of equations:

A+C = 0

B+D = 1

4A+2C+E = 0

4B+2D+F = 1

4A = 0

4B = 4

Solving the system of equations results with:

A = 0

C = 0

E = 0

B = 1

D = 0

F = -3

Thus, our original integrand can be converted into the following partial fraction decomposition:

(x⁴+x²+4)/(x²(x²+2)²) = 1/x² - 3/(x²+2)²

That's the first difficult part. Now we integrate the right side, which has one easy piece, and one hard piece.

The integral of 1/x² is just -1/x.

The integral of - 3/(x²+2)² will require trigonometric substitution. I can't draw a triangle here to show you the sides, but I will write out the substitutions I used:

For ease, going forward, I will use T to stand theta, and sqrt() to represent the square root of an expression.

sqrt(2)sec(T) = sqrt(x²+2)

sqrt(2)tan(T) = x

sqrt(2)sec²(T)dT = dx

Substituting those into our integral, we should end up with -3sqrt(2)/4 (1/sec²(T))dT

1/sec²(T) is simply cos^2(T), and to integrate cos²(T) we power reduce into 1/2(1+cos(2T))

Finally able to integrate, we end up with:

-3sqrt(2)/8 (T + 1/2sin(2T)) Utilizing the double angle formula for sine, we simplify to:

-3sqrt(2)/8 (T + sin(T)cos(T))

In the end we can use our trig substitutions and triangle to substitute back in for T:

-3sqrt(2)/8 (arctan(x/sqrt(2)) + sqrt(2)x/(x²+2)) + C

That was a good one! Hope this helps.