Problem to solve
Find the indefinite integral \[\int\frac{3\sqrt[3]{2x-1}-4}{\sqrt[6]{2x-1}-1} dx\]
Solution
STEP 1
Take the integral: integral(3 (2 x - 1)^(1/3) - 4)/((2 x - 1)^(1/6) - 1) dx
STEP 2
For the integrand (3 (2 x - 1)^(1/3) - 4)/((2 x - 1)^(1/6) - 1), substitute u = 2 x - 1 and du = 2 dx: = 1/2 integral(3 u^(1/3) - 4)/(u^(1/6) - 1) du
STEP 3
For the integrand (3 u^(1/3) - 4)/(u^(1/6) - 1), substitute s = u^(1/6) and ds = 1/(6 u^(5/6)) du: = 3 integral(3 s^7 - 4 s^5)/(s - 1) ds
STEP 4
For the integrand (3 s^7 - 4 s^5)/(s - 1), do long division: = 3 integral3 s^6 + 3 s^5 - s^4 - s^3 - s^2 - s - 1/(s - 1) - 1 ds
STEP 5
Integrate the sum term by term and factor out constants: = 9 integral s^6 ds + 9 integral s^5 ds - 3 integral s^4 ds - 3 integral s^3 ds - 3 integral s^2 ds - 3 integral s ds - 3 integral1/(s - 1) ds - 3 integral1 ds
STEP 6
The integral of s^6 is s^7/7: = (9 s^7)/7 + 9 integral s^5 ds - 3 integral s^4 ds - 3 integral s^3 ds - 3 integral s^2 ds - 3 integral s ds - 3 integral1/(s - 1) ds - 3 integral1 ds
STEP 7
The integral of s^5 is s^6/6: = (3 s^6)/2 + (9 s^7)/7 - 3 integral s^4 ds - 3 integral s^3 ds - 3 integral s^2 ds - 3 integral s ds - 3 integral1/(s - 1) ds - 3 integral1 ds
STEP 8
The integral of s^4 is s^5/5: = -(3 s^5)/5 + (3 s^6)/2 + (9 s^7)/7 - 3 integral s^3 ds - 3 integral s^2 ds - 3 integral s ds - 3 integral1/(s - 1) ds - 3 integral1 ds
STEP 9
The integral of s^3 is s^4/4: = -(3 s^4)/4 - (3 s^5)/5 + (3 s^6)/2 + (9 s^7)/7 - 3 integral s^2 ds - 3 integral s ds - 3 integral1/(s - 1) ds - 3 integral1 ds
STEP 10
The integral of s^2 is s^3/3: = -s^3 - (3 s^4)/4 - (3 s^5)/5 + (3 s^6)/2 + (9 s^7)/7 - 3 integral s ds - 3 integral1/(s - 1) ds - 3 integral1 ds
STEP 11
The integral of s is s^2/2: = -(3 s^2)/2 - s^3 - (3 s^4)/4 - (3 s^5)/5 + (3 s^6)/2 + (9 s^7)/7 - 3 integral1/(s - 1) ds - 3 integral1 ds
STEP 12
For the integrand 1/(s - 1), substitute p = s - 1 and dp = ds: = -(3 s^2)/2 - s^3 - (3 s^4)/4 - (3 s^5)/5 + (3 s^6)/2 + (9 s^7)/7 - 3 integral1/p dp - 3 integral1 ds
STEP 13
The integral of 1/p is log(p): = -(3 s^2)/2 - s^3 - (3 s^4)/4 - (3 s^5)/5 + (3 s^6)/2 + (9 s^7)/7 - 3 log(p) - 3 integral1 ds
STEP 14
The integral of 1 is s: = -3 log(p) + (9 s^7)/7 + (3 s^6)/2 - (3 s^5)/5 - (3 s^4)/4 - s^3 - (3 s^2)/2 - 3 s + constant
STEP 15
Substitute back for p = s - 1: = (9 s^7)/7 + (3 s^6)/2 - (3 s^5)/5 - (3 s^4)/4 - s^3 - (3 s^2)/2 - 3 s - 3 log(s - 1) + constant
STEP 16
Substitute back for s = u^(1/6): = (9 u^(7/6))/7 - (3 u^(5/6))/5 - (3 u^(2/3))/4 + (3 u)/2 - sqrt(u) - (3 u^(1/3))/2 - 3 u^(1/6) - 3 log(u^(1/6) - 1) + constant
STEP 17
Substitute back for u = 2 x - 1: = 9/7 (2 x - 1)^(7/6) - 3/5 (2 x - 1)^(5/6) - 3/4 (2 x - 1)^(2/3) - sqrt(2 x - 1) - 3/2 (2 x - 1)^(1/3) - 3 (2 x - 1)^(1/6) + 3 x - 3 log((2 x - 1)^(1/6) - 1) - 3/2 + constant
STEP 18
Factor the answer a different way: = 3/7 x (6 (2 x - 1)^(1/6) + 7) - 3/5 (2 x - 1)^(5/6) - 3/4 (2 x - 1)^(2/3) - sqrt(2 x - 1) - 3/2 (2 x - 1)^(1/3) - 30/7 (2 x - 1)^(1/6) - 3 log((2 x - 1)^(1/6) - 1) - 3/2 + constant
STEP 19
Which is equivalent for restricted x values to: Answer: | | = 3 (3/7 (2 x - 1)^(7/6) - 1/5 (2 x - 1)^(5/6) - 1/4 (2 x - 1)^(2/3) - 1/3 sqrt(2 x - 1) - 1/2 (2 x - 1)^(1/3) - (2 x - 1)^(1/6) + x - log(1 - (2 x - 1)^(1/6))) + constant
More problems
Number
001Find the indefinite integral \(\int\frac{x^2+2}{x^3+5x^2+2x-8} dx\)
002Find the indefinite integral \(\int\frac{4x^4-2}{x^3+4x^2+x-6} dx\)
003Find the indefinite integral \(\int\frac{4x^4-22x^3+11x^2-99x-34}{x^3-6x^2+4x-24} dx\)
004Find the indefinite integral \(\int\frac{3x^4+4x^2+5x}{(x-1)(x^2+1)^2} dx\)
005Find the indefinite integral \(\int\frac{x^6+6x^4+8x^2-4}{(x^2+4)(x^2+3)} dx\)
006Find the indefinite integral \(\int(-4x^4+6x^2+8x-1)\cos(x) dx\)
007Find the indefinite integral \(\int(-7x^5+2x^4-2)\cdot3^xdx\)
008Find the indefinite integral \(\int\arcsin^4(x) dx\)
009Find the indefinite integral \(\int\sin(\ln(x)) dx\)
010Find the indefinite integral \(\int(12x^{15}-3x^7-4x^3)\cdot e^{2x^4} dx\)
011Find the indefinite integral \(\int\frac{\sqrt{(3x^2+4)^3}}{x^4} dx\)
012Find the indefinite integral \(\int\frac{x^2}{\sqrt{2x^2-5}} dx\)
013Find the indefinite integral \(\int\frac{x^6}{\sqrt{(x^2+4)^3}} dx\)
014Find the indefinite integral \(\int x^2\sqrt{4-3x^2} dx\)
015Find the indefinite integral \(\int x^4\cdot\sqrt{x^2+4x+5} dx\)
016Find the indefinite integral \(\int\frac{\sqrt{x+2}}{x+1} dx\)
017Find the indefinite integral \(\int\frac{\sqrt{x+2}-4}{5-\sqrt{x+2}} dx\)
018Find the indefinite integral \(\int\frac{3\sqrt[3]{2x-1}-4}{\sqrt[6]{2x-1}-1} dx\)
019Find the indefinite integral \(\int\sqrt{\frac{x+3}{1-x}} dx\)
020Find the indefinite integral \(\int\frac{\sqrt[3]{x+1}-\sqrt{x+1}}{x} dx\)
021Find the indefinite integral \(\int\frac{\sin(x)-3\cos(x)}{\cos(x)-2} dx\)
022Find the indefinite integral \(\int\frac{3\sec(x)-\cot(x)}{1-\csc(x)} dx\)
023Find the indefinite integral \(\int\frac{\tan^2(x)+2}{\tan(x)+2} dx\)
024Find the indefinite integral \(\int\frac{3\sin^4(x)-2\cos^2(x)}{1+\sin^2(x)} dx\)
025Find the indefinite integral \(\int\tan^4(x)\sec^5(x) dx\)
Back
Edited by tododemate® - 2021