Problem to solve
Find the indefinite integral \[\int\frac{4x^4-2}{x^3+4x^2+x-6} dx\]
Solution
STEP 1
Take the integral: integral(4 x^4 - 2)/(x^3 + 4 x^2 + x - 6) dx
STEP 2
For the integrand (4 x^4 - 2)/(x^3 + 4 x^2 + x - 6), do long division: = integral4 x + 1/(6 (x - 1)) - 62/(3 (x + 2)) + 161/(2 (x + 3)) - 16 dx
STEP 3
Integrate the sum term by term and factor out constants: = 161/2 integral1/(x + 3) dx - 62/3 integral1/(x + 2) dx + 4 integral x dx + 1/6 integral1/(x - 1) dx - 16 integral1 dx
STEP 4
For the integrand 1/(x + 3), substitute u = x + 3 and du = dx: = 161/2 integral1/u du - 62/3 integral1/(x + 2) dx + 4 integral x dx + 1/6 integral1/(x - 1) dx - 16 integral1 dx
STEP 5
The integral of 1/u is log(u): = (161 log(u))/2 - 62/3 integral1/(x + 2) dx + 4 integral x dx + 1/6 integral1/(x - 1) dx - 16 integral1 dx
STEP 6
For the integrand 1/(x + 2), substitute s = x + 2 and ds = dx: = (161 log(u))/2 - 62/3 integral1/s ds + 4 integral x dx + 1/6 integral1/(x - 1) dx - 16 integral1 dx
STEP 7
The integral of 1/s is log(s): = -(62 log(s))/3 + (161 log(u))/2 + 4 integral x dx + 1/6 integral1/(x - 1) dx - 16 integral1 dx
STEP 8
The integral of x is x^2/2: = 2 x^2 - (62 log(s))/3 + (161 log(u))/2 + 1/6 integral1/(x - 1) dx - 16 integral1 dx
STEP 9
For the integrand 1/(x - 1), substitute p = x - 1 and dp = dx: = 2 x^2 - (62 log(s))/3 + (161 log(u))/2 + 1/6 integral1/p dp - 16 integral1 dx
STEP 10
The integral of 1/p is log(p): = 2 x^2 + log(p)/6 - (62 log(s))/3 + (161 log(u))/2 - 16 integral1 dx
STEP 11
The integral of 1 is x: = log(p)/6 - (62 log(s))/3 + (161 log(u))/2 + 2 x^2 - 16 x + constant
STEP 12
Substitute back for p = x - 1: = -(62 log(s))/3 + (161 log(u))/2 + 2 x^2 - 16 x + 1/6 log(x - 1) + constant
STEP 13
Substitute back for s = x + 2: = (161 log(u))/2 + 2 x^2 - 16 x + 1/6 log(x - 1) - 62/3 log(x + 2) + constant
STEP 14
Substitute back for u = x + 3: = 2 x^2 - 16 x + 1/6 log(x - 1) - 62/3 log(x + 2) + 161/2 log(x + 3) + constant
STEP 15
Factor the answer a different way: = 1/6 (12 x^2 - 96 x + log(x - 1) - 124 log(x + 2) + 483 log(x + 3)) + constant
STEP 16
Which is equivalent for restricted x values to: Answer: | | = 1/6 (12 x^2 - 96 x + log(1 - x) - 124 log(x + 2) + 483 log(x + 3)) + constant
More problems
Number
001Find the indefinite integral \(\int\frac{x^2+2}{x^3+5x^2+2x-8} dx\)
002Find the indefinite integral \(\int\frac{4x^4-2}{x^3+4x^2+x-6} dx\)
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Edited by tododemate® - 2021