Problem to solve
Find the indefinite integral \[\int\frac{3x^4+4x^2+5x}{[x-1][x^2+1]^2} dx\]
Solution
STEP 1
Take the integral: integral(3 x^4 + 4 x^2 + 5 x)/((x - 1) (x^2 + 1)^2) dx
STEP 2
For the integrand (3 x^4 + 4 x^2 + 5 x)/((x - 1) (x^2 + 1)^2), use partial fractions: = integral(3 - 2 x)/(x^2 + 1)^2 + 3/(x - 1) dx
STEP 3
Integrate the sum term by term and factor out constants: = integral(3 - 2 x)/(x^2 + 1)^2 dx + 3 integral1/(x - 1) dx
STEP 4
For the integrand (3 - 2 x)/(x^2 + 1)^2, substitute x = tan(u) and dx = sec^2(u) du. Then (x^2 + 1)^2 = (tan^2(u) + 1)^2 = sec^4(u) and u = tan^(-1)(x): = integral cos^2(u) (3 - 2 tan(u)) du + 3 integral1/(x - 1) dx
STEP 5
Write cos^2(u) as 1 - sin^2(u): = integral(1 - sin^2(u)) (3 - 2 tan(u)) du + 3 integral1/(x - 1) dx
STEP 6
Expanding the integrand (1 - sin^2(u)) (3 - 2 tan(u)) gives -3 sin^2(u) - 2 tan(u) + 2 sin^2(u) tan(u) + 3: = integral(-3 sin^2(u) - 2 tan(u) + 2 sin^2(u) tan(u) + 3) du + 3 integral1/(x - 1) dx
STEP 7
Integrate the sum term by term and factor out constants: = 2 integral sin^2(u) tan(u) du - 2 integral tan(u) du - 3 integral sin^2(u) du + 3 integral1 du + 3 integral1/(x - 1) dx
STEP 8
Write sin^2(u) as 1 - cos^2(u): = 2 integral(1 - cos^2(u)) tan(u) du - 2 integral tan(u) du - 3 integral sin^2(u) du + 3 integral1 du + 3 integral1/(x - 1) dx
STEP 9
Expanding the integrand (1 - cos^2(u)) tan(u) gives tan(u) - sin(u) cos(u): = 2 integral(tan(u) - sin(u) cos(u)) du - 2 integral tan(u) du - 3 integral sin^2(u) du + 3 integral1 du + 3 integral1/(x - 1) dx
STEP 10
Integrate the sum term by term and factor out constants: = -2 integral sin(u) cos(u) du - 3 integral sin^2(u) du + 3 integral1 du + 3 integral1/(x - 1) dx
STEP 11
For the integrand sin(u) cos(u), substitute s = cos(u) and ds = -sin(u) du: = 2 integral s ds - 3 integral sin^2(u) du + 3 integral1 du + 3 integral1/(x - 1) dx
STEP 12
The integral of s is s^2/2: = s^2 - 3 integral sin^2(u) du + 3 integral1 du + 3 integral1/(x - 1) dx
STEP 13
Write sin^2(u) as 1/2 - 1/2 cos(2 u): = s^2 - 3 integral(1/2 - 1/2 cos(2 u)) du + 3 integral1 du + 3 integral1/(x - 1) dx
STEP 14
Integrate the sum term by term and factor out constants: = s^2 + 3/2 integral cos(2 u) du + 3/2 integral1 du + 3 integral1/(x - 1) dx
STEP 15
For the integrand cos(2 u), substitute p = 2 u and dp = 2 du: = s^2 + 3/4 integral cos(p) dp + 3/2 integral1 du + 3 integral1/(x - 1) dx
STEP 16
The integral of cos(p) is sin(p): = s^2 + (3 sin(p))/4 + 3/2 integral1 du + 3 integral1/(x - 1) dx
STEP 17
The integral of 1 is u: = s^2 + (3 u)/2 + (3 sin(p))/4 + 3 integral1/(x - 1) dx
STEP 18
For the integrand 1/(x - 1), substitute w = x - 1 and dw = dx: = s^2 + (3 u)/2 + (3 sin(p))/4 + 3 integral1/w dw
STEP 19
The integral of 1/w is log(w): = (3 sin(p))/4 + s^2 + (3 u)/2 + 3 log(w) + constant
STEP 20
Substitute back for w = x - 1: = (3 sin(p))/4 + s^2 + (3 u)/2 + 3 log(x - 1) + constant
STEP 21
Substitute back for p = 2 u: = s^2 + (3 u)/2 + 3/4 sin(2 u) + 3 log(x - 1) + constant
STEP 22
Substitute back for s = cos(u): = (3 u)/2 + 3/4 sin(2 u) + cos^2(u) + 3 log(x - 1) + constant
STEP 23
Substitute back for u = tan^(-1)(x): = 3 log(x - 1) + 3/2 tan^(-1)(x) + 3/4 sin(2 tan^(-1)(x)) + cos(tan^(-1)(x))^2 + constant
STEP 24
Simplify using cos(tan^(-1)(z)) = 1/sqrt(z^2 + 1): = (3 x + 2)/(2 x^2 + 2) + 3 log(x - 1) + 3/2 tan^(-1)(x) + constant
STEP 25
Which is equivalent for restricted x values to: Answer: | | = 1/2 ((3 x + 2)/(x^2 + 1) + 6 log(1 - x) + 3 tan^(-1)(x)) + constant
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Edited by tododemate® - 2021