Problem to solve
Find the indefinite integral \[\int\frac{x^6+6x^4+8x^2-4}{[x^2+4][x^2+3]} dx\]
Solution
STEP 1
Take the integral: integral(x^6 + 6 x^4 + 8 x^2 - 4)/((x^2 + 3) (x^2 + 4)) dx
STEP 2
For the integrand (x^6 + 6 x^4 + 8 x^2 - 4)/((x^2 + 3) (x^2 + 4)), do long division: = integral x^2 - 1/(x^2 + 3) + 4/(x^2 + 4) - 1 dx
STEP 3
Integrate the sum term by term and factor out constants: = 4 integral1/(x^2 + 4) dx - integral1/(x^2 + 3) dx + integral x^2 dx - integral1 dx
STEP 4
Factor 4 from the denominator: = 4 integral1/(4 (x^2/4 + 1)) dx - integral1/(x^2 + 3) dx + integral x^2 dx - integral1 dx
STEP 5
Factor out constants: = integral1/(x^2/4 + 1) dx - integral1/(x^2 + 3) dx + integral x^2 dx - integral1 dx
STEP 6
For the integrand 1/(x^2/4 + 1), substitute u = x/2 and du = 1/2 dx: = 2 integral1/(u^2 + 1) du - integral1/(x^2 + 3) dx + integral x^2 dx - integral1 dx
STEP 7
The integral of 1/(u^2 + 1) is tan^(-1)(u): = 2 tan^(-1)(u) - integral1/(x^2 + 3) dx + integral x^2 dx - integral1 dx
STEP 8
Factor 3 from the denominator: = 2 tan^(-1)(u) - integral1/(3 (x^2/3 + 1)) dx + integral x^2 dx - integral1 dx
STEP 9
Factor out constants: = 2 tan^(-1)(u) - 1/3 integral1/(x^2/3 + 1) dx + integral x^2 dx - integral1 dx
STEP 10
For the integrand 1/(x^2/3 + 1), substitute s = x/sqrt(3) and ds = 1/sqrt(3) dx: = 2 tan^(-1)(u) - 1/sqrt(3) integral1/(s^2 + 1) ds + integral x^2 dx - integral1 dx
STEP 11
The integral of 1/(s^2 + 1) is tan^(-1)(s): = -(tan^(-1)(s))/sqrt(3) + 2 tan^(-1)(u) + integral x^2 dx - integral1 dx
STEP 12
The integral of x^2 is x^3/3: = x^3/3 - (tan^(-1)(s))/sqrt(3) + 2 tan^(-1)(u) - integral1 dx
STEP 13
The integral of 1 is x: = -(tan^(-1)(s))/sqrt(3) + 2 tan^(-1)(u) + x^3/3 - x + constant
STEP 14
Substitute back for s = x/sqrt(3): = 2 tan^(-1)(u) + x^3/3 - x - (tan^(-1)(x/sqrt(3)))/sqrt(3) + constant
STEP 15
Substitute back for u = x/2: Answer: | | = x^3/3 - x + 2 tan^(-1)(x/2) - (tan^(-1)(x/sqrt(3)))/sqrt(3) + constant
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Edited by tododemate® - 2021