Problem to solve
Find the indefinite integral \[\int\frac{\sqrt[3]{x+1}-\sqrt{x+1}}{x} dx\]
Solution
STEP 1
Take the integral: integral((x + 1)^(1/3) - sqrt(x + 1))/x dx
STEP 2
For the integrand ((x + 1)^(1/3) - sqrt(x + 1))/x, substitute u = sqrt(x + 1) and du = 1/(2 sqrt(x + 1)) dx: = 2 integral(u ((u^2)^(1/3) - u))/(u^2 - 1) du
STEP 3
Expanding the integrand (u ((u^2)^(1/3) - u))/(u^2 - 1) gives (u (u^2)^(1/3))/(u^2 - 1) - u^2/(u^2 - 1): = 2 integral((u (u^2)^(1/3))/(u^2 - 1) - u^2/(u^2 - 1)) du
STEP 4
Integrate the sum term by term and factor out constants: = 2 integral(u (u^2)^(1/3))/(u^2 - 1) du - 2 integral u^2/(u^2 - 1) du
STEP 5
For the integrand (u (u^2)^(1/3))/(u^2 - 1), substitute s = u^2 and ds = 2 u du: = integral s^(1/3)/(s - 1) ds - 2 integral u^2/(u^2 - 1) du
STEP 6
For the integrand s^(1/3)/(s - 1), substitute p = s^(1/3) and dp = 1/(3 s^(2/3)) ds: = 3 integral p^3/(p^3 - 1) dp - 2 integral u^2/(u^2 - 1) du
STEP 7
For the integrand p^3/(p^3 - 1), do long division: = 3 integral(-p - 2)/(3 (p^2 + p + 1)) + 1/(3 (p - 1)) + 1 dp - 2 integral u^2/(u^2 - 1) du
STEP 8
Integrate the sum term by term and factor out constants: = integral(-p - 2)/(p^2 + p + 1) dp + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 9
Rewrite the integrand (-p - 2)/(p^2 + p + 1) as -(2 p + 1)/(2 (p^2 + p + 1)) - 3/(2 (p^2 + p + 1)): = integral(-(2 p + 1)/(2 (p^2 + p + 1)) - 3/(2 (p^2 + p + 1))) dp + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 10
Integrate the sum term by term and factor out constants: = -1/2 integral(2 p + 1)/(p^2 + p + 1) dp - 3/2 integral1/(p^2 + p + 1) dp + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 11
For the integrand (2 p + 1)/(p^2 + p + 1), substitute w = p^2 + p + 1 and dw = (2 p + 1) dp: = -1/2 integral1/w dw - 3/2 integral1/(p^2 + p + 1) dp + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 12
The integral of 1/w is log(w): = -log(w)/2 - 3/2 integral1/(p^2 + p + 1) dp + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 13
For the integrand 1/(p^2 + p + 1), complete the square: = -log(w)/2 - 3/2 integral1/((p + 1/2)^2 + 3/4) dp + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 14
For the integrand 1/((p + 1/2)^2 + 3/4), substitute v = p + 1/2 and dv = dp: = -log(w)/2 - 3/2 integral1/(v^2 + 3/4) dv + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 15
Factor 3/4 from the denominator: = -log(w)/2 - 3/2 integral4/(3 ((4 v^2)/3 + 1)) dv + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 16
Factor out constants: = -log(w)/2 - 2 integral1/((4 v^2)/3 + 1) dv + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 17
For the integrand 1/((4 v^2)/3 + 1), substitute z_1 = (2 v)/sqrt(3) and dz_1 = 2/sqrt(3) dv: = -log(w)/2 - sqrt(3) integral1/(z_1^2 + 1) dz_1 + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 18
The integral of 1/(z_1^2 + 1) is tan^(-1)(z_1): = -sqrt(3) tan^(-1)(z_1) - log(w)/2 + integral1/(p - 1) dp + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 19
For the integrand 1/(p - 1), substitute z_2 = p - 1 and dz_2 = dp: = -sqrt(3) tan^(-1)(z_1) - log(w)/2 + integral1/z_2 dz_2 + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 20
The integral of 1/z_2 is log(z_2): = -sqrt(3) tan^(-1)(z_1) - log(w)/2 + log(z_2) + 3 integral1 dp - 2 integral u^2/(u^2 - 1) du
STEP 21
The integral of 1 is p: = 3 p - sqrt(3) tan^(-1)(z_1) - log(w)/2 + log(z_2) - 2 integral u^2/(u^2 - 1) du
STEP 22
For the integrand u^2/(u^2 - 1), do long division: = 3 p - sqrt(3) tan^(-1)(z_1) - log(w)/2 + log(z_2) - 2 integral-1/(2 (u + 1)) + 1 + 1/(2 (u - 1)) du
STEP 23
Integrate the sum term by term and factor out constants: = 3 p - sqrt(3) tan^(-1)(z_1) - log(w)/2 + log(z_2) + integral1/(u + 1) du - integral1/(u - 1) du - 2 integral1 du
STEP 24
For the integrand 1/(u + 1), substitute z_3 = u + 1 and dz_3 = du: = 3 p - sqrt(3) tan^(-1)(z_1) - log(w)/2 + log(z_2) + integral1/z_3 dz_3 - integral1/(u - 1) du - 2 integral1 du
STEP 25
The integral of 1/z_3 is log(z_3): = 3 p - sqrt(3) tan^(-1)(z_1) - log(w)/2 + log(z_2) + log(z_3) - integral1/(u - 1) du - 2 integral1 du
STEP 26
For the integrand 1/(u - 1), substitute z_4 = u - 1 and dz_4 = du: = 3 p - sqrt(3) tan^(-1)(z_1) - log(w)/2 + log(z_2) + log(z_3) - integral1/z_4 dz_4 - 2 integral1 du
STEP 27
The integral of 1/z_4 is log(z_4): = 3 p - sqrt(3) tan^(-1)(z_1) - log(w)/2 + log(z_2) + log(z_3) - log(z_4) - 2 integral1 du
STEP 28
The integral of 1 is u: = 3 p - 2 u - log(w)/2 + log(z_2) + log(z_3) - log(z_4) - sqrt(3) tan^(-1)(z_1) + constant
STEP 29
Substitute back for z_4 = u - 1: = 3 p - 2 u - log(u - 1) - log(w)/2 + log(z_2) + log(z_3) - sqrt(3) tan^(-1)(z_1) + constant
STEP 30
Substitute back for z_3 = u + 1: = 3 p - 2 u - log(u - 1) + log(u + 1) - log(w)/2 + log(z_2) - sqrt(3) tan^(-1)(z_1) + constant
STEP 31
Substitute back for z_2 = p - 1: = 3 p + log(p - 1) - 2 u - log(u - 1) + log(u + 1) - log(w)/2 - sqrt(3) tan^(-1)(z_1) + constant
STEP 32
Substitute back for z_1 = (2 v)/sqrt(3): = 3 p + log(p - 1) - 2 u - log(u - 1) + log(u + 1) - sqrt(3) tan^(-1)((2 v)/sqrt(3)) - log(w)/2 + constant
STEP 33
Substitute back for v = p + 1/2: = 3 p + log(p - 1) - sqrt(3) tan^(-1)((2 p + 1)/sqrt(3)) - 2 u - log(u - 1) + log(u + 1) - log(w)/2 + constant
STEP 34
Substitute back for w = p^2 + p + 1: = -1/2 log(p^2 + p + 1) + 3 p + log(p - 1) - sqrt(3) tan^(-1)((2 p + 1)/sqrt(3)) - 2 u - log(u - 1) + log(u + 1) + constant
STEP 35
Substitute back for p = s^(1/3): = -1/2 log(s^(2/3) + s^(1/3) + 1) + 3 s^(1/3) + log(s^(1/3) - 1) - sqrt(3) tan^(-1)((2 s^(1/3) + 1)/sqrt(3)) - 2 u - log(u - 1) + log(u + 1) + constant
STEP 36
Substitute back for s = u^2: = 3 (u^2)^(1/3) + log((u^2)^(1/3) - 1) - 1/2 log((u^2)^(2/3) + (u^2)^(1/3) + 1) - sqrt(3) tan^(-1)((2 (u^2)^(1/3) + 1)/sqrt(3)) - 2 u - log(u - 1) + log(u + 1) + constant
STEP 37
Substitute back for u = sqrt(x + 1): = -2 sqrt(x + 1) + 3 (x + 1)^(1/3) + log((x + 1)^(1/3) - 1) - log(sqrt(x + 1) - 1) + log(sqrt(x + 1) + 1) - 1/2 log((x + 1)^(2/3) + (x + 1)^(1/3) + 1) - sqrt(3) tan^(-1)((2 (x + 1)^(1/3) + 1)/sqrt(3)) + constant
STEP 38
An alternative form of the integral is: = -2 sqrt(x + 1) + 3 (x + 1)^(1/3) + log((((x + 1)^(1/3) - 1) (sqrt(x + 1) + 1))/(sqrt(x + 1) - 1)) - 1/2 log((x + 1)^(2/3) + (x + 1)^(1/3) + 1) - sqrt(3) tan^(-1)((2 (x + 1)^(1/3) + 1)/sqrt(3)) + constant
STEP 39
Which is equivalent for restricted x values to: Answer: | | = 1/2 (-4 sqrt(x + 1) + 6 (x + 1)^(1/3) + 4 log((x + 1)^(1/6) + 1) + log((x + 1)^(1/3) - (x + 1)^(1/6) + 1) - 3 log((x + 1)^(1/3) + (x + 1)^(1/6) + 1) + 2 sqrt(3) tan^(-1)((1 - 2 (x + 1)^(1/6))/sqrt(3)) + 2 sqrt(3) tan^(-1)((2 (x + 1)^(1/6) + 1)/sqrt(3))) + constant
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Edited by tododemate® - 2021