Problem to solve
Find the indefinite integral \[\int\frac{4x^4-22x^3+11x^2-99x-34}{x^3-6x^2+4x-24} dx\]
Solution
STEP 1
Take the integral: integral(4 x^4 - 22 x^3 + 11 x^2 - 99 x - 34)/(x^3 - 6 x^2 + 4 x - 24) dx
STEP 2
For the integrand (4 x^4 - 22 x^3 + 11 x^2 - 99 x - 34)/(x^3 - 6 x^2 + 4 x - 24), do long division: = integral(2 x + 1)/(x^2 + 4) + 4 x + 5/(x - 6) + 2 dx
STEP 3
Integrate the sum term by term and factor out constants: = integral(2 x + 1)/(x^2 + 4) dx + 4 integral x dx + 5 integral1/(x - 6) dx + 2 integral1 dx
STEP 4
Expanding the integrand (2 x + 1)/(x^2 + 4) gives (2 x)/(x^2 + 4) + 1/(x^2 + 4): = integral((2 x)/(x^2 + 4) + 1/(x^2 + 4)) dx + 4 integral x dx + 5 integral1/(x - 6) dx + 2 integral1 dx
STEP 5
Integrate the sum term by term and factor out constants: = 2 integral x/(x^2 + 4) dx + integral1/(x^2 + 4) dx + 4 integral x dx + 5 integral1/(x - 6) dx + 2 integral1 dx
STEP 6
For the integrand x/(x^2 + 4), substitute u = x^2 + 4 and du = 2 x dx: = integral1/u du + integral1/(x^2 + 4) dx + 4 integral x dx + 5 integral1/(x - 6) dx + 2 integral1 dx
STEP 7
The integral of 1/u is log(u): = log(u) + integral1/(x^2 + 4) dx + 4 integral x dx + 5 integral1/(x - 6) dx + 2 integral1 dx
STEP 8
Factor 4 from the denominator: = log(u) + integral1/(4 (x^2/4 + 1)) dx + 4 integral x dx + 5 integral1/(x - 6) dx + 2 integral1 dx
STEP 9
Factor out constants: = log(u) + 1/4 integral1/(x^2/4 + 1) dx + 4 integral x dx + 5 integral1/(x - 6) dx + 2 integral1 dx
STEP 10
For the integrand 1/(x^2/4 + 1), substitute s = x/2 and ds = 1/2 dx: = log(u) + 1/2 integral1/(s^2 + 1) ds + 4 integral x dx + 5 integral1/(x - 6) dx + 2 integral1 dx
STEP 11
The integral of 1/(s^2 + 1) is tan^(-1)(s): = 1/2 tan^(-1)(s) + log(u) + 4 integral x dx + 5 integral1/(x - 6) dx + 2 integral1 dx
STEP 12
The integral of x is x^2/2: = 2 x^2 + 1/2 tan^(-1)(s) + log(u) + 5 integral1/(x - 6) dx + 2 integral1 dx
STEP 13
For the integrand 1/(x - 6), substitute p = x - 6 and dp = dx: = 2 x^2 + 1/2 tan^(-1)(s) + log(u) + 5 integral1/p dp + 2 integral1 dx
STEP 14
The integral of 1/p is log(p): = 2 x^2 + 1/2 tan^(-1)(s) + 5 log(p) + log(u) + 2 integral1 dx
STEP 15
The integral of 1 is x: = 5 log(p) + 1/2 tan^(-1)(s) + log(u) + 2 x^2 + 2 x + constant
STEP 16
Substitute back for p = x - 6: = 1/2 tan^(-1)(s) + log(u) + 2 x^2 + 2 x + 5 log(x - 6) + constant
STEP 17
Substitute back for s = x/2: = log(u) + 2 x^2 + 2 x + 5 log(x - 6) + 1/2 tan^(-1)(x/2) + constant
STEP 18
Substitute back for u = x^2 + 4: = 2 x^2 + log(x^2 + 4) + 2 x + 5 log(x - 6) + 1/2 tan^(-1)(x/2) + constant
STEP 19
Which is equivalent for restricted x values to: Answer: | | = 2 x^2 + log(x^2 + 4) + 2 x + 5 log(6 - x) + 1/2 tan^(-1)(x/2) + constant
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Edited by tododemate® - 2021