I'm having trouble solving finding the integral $int (sqrt[3]x(x-1)) dx$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Integral question - $intfracsinsqrtxsqrtx$I'm having problems solving this indefinite integralTrouble solving $intsqrt1-x^2 , dx$How to solve $intsqrt1-x^8dx$Evaluate indefinite integral. $intsqrt225-t^2dt$Solve integral $int sqrt3x^2 - 2x dx$Evaluating the indefinite integral $intfracsqrt1-xxsqrt1+x,dx$Find the indefinite integral: $int sqrtx+1 over sqrtx+2 - sqrtx-2 dx$Evaluate the algebraic indefinite integral $int fracsqrtx+1-sqrtx-1sqrtx+1+sqrtx-1dx$The integral $int sqrt x sqrt[3] x sqrt[4] x sqrt[5] x cdots dx$
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I'm having trouble solving finding the integral $int (sqrt[3]x(x-1)) dx$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Integral question - $intfracsinsqrtxsqrtx$I'm having problems solving this indefinite integralTrouble solving $intsqrt1-x^2 , dx$How to solve $intsqrt1-x^8dx$Evaluate indefinite integral. $intsqrt225-t^2dt$Solve integral $int sqrt3x^2 - 2x dx$Evaluating the indefinite integral $intfracsqrt1-xxsqrt1+x,dx$Find the indefinite integral: $int sqrtx+1 over sqrtx+2 - sqrtx-2 dx$Evaluate the algebraic indefinite integral $int fracsqrtx+1-sqrtx-1sqrtx+1+sqrtx-1dx$The integral $int sqrt x sqrt[3] x sqrt[4] x sqrt[5] x cdots dx$
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$$int (sqrt[3]x(x-1)) dx$$
I think I can start off by making x-1 v'x and cube root x ux...?
calculus integration
edited Apr 1 at 16:02
Eevee Trainer
10.6k31842
asked Apr 1 at 6:18
ssgg1ssgg1
23
$endgroup$
add a comment |
$begingroup$
$$int (sqrt[3]x(x-1)) dx$$
I think I can start off by making x-1 v'x and cube root x ux...?
calculus integration
edited Apr 1 at 16:02
Eevee Trainer
10.6k31842
asked Apr 1 at 6:18
ssgg1ssgg1
23
$endgroup$
$begingroup$
For clarity, are you finding $$int sqrt[3]x(x-1) dx$$ By the way, here's a reference for MathJax you might find useful, which is preferred for writing up and rendering your math text on this site.
$endgroup$
– Eevee Trainer
Apr 1 at 6:22
$begingroup$
Have you tried $x=sec^2 t$?
$endgroup$
– J.G.
Apr 1 at 6:35
$begingroup$
I fixed the equation in the question!
$endgroup$
– ssgg1
Apr 1 at 13:16
add a comment |
$begingroup$
$$int (sqrt[3]x(x-1)) dx$$
I think I can start off by making x-1 v'x and cube root x ux...?
calculus integration
edited Apr 1 at 16:02
Eevee Trainer
10.6k31842
asked Apr 1 at 6:18
ssgg1ssgg1
23
$endgroup$
$$int (sqrt[3]x(x-1)) dx$$
I think I can start off by making x-1 v'x and cube root x ux...?
calculus integration
calculus integration
edited Apr 1 at 16:02
Eevee Trainer
10.6k31842
asked Apr 1 at 6:18
ssgg1ssgg1
23
edited Apr 1 at 16:02
Eevee Trainer
10.6k31842
asked Apr 1 at 6:18
ssgg1ssgg1
23
edited Apr 1 at 16:02
Eevee Trainer
10.6k31842
edited Apr 1 at 16:02
Eevee Trainer
10.6k31842
edited Apr 1 at 16:02
Eevee Trainer
10.6k31842
10.6k31842
asked Apr 1 at 6:18
ssgg1ssgg1
23
asked Apr 1 at 6:18
ssgg1ssgg1
23
asked Apr 1 at 6:18
ssgg1ssgg1
23
23
$begingroup$
For clarity, are you finding $$int sqrt[3]x(x-1) dx$$ By the way, here's a reference for MathJax you might find useful, which is preferred for writing up and rendering your math text on this site.
$endgroup$
– Eevee Trainer
Apr 1 at 6:22
$begingroup$
Have you tried $x=sec^2 t$?
$endgroup$
– J.G.
Apr 1 at 6:35
$begingroup$
I fixed the equation in the question!
$endgroup$
– ssgg1
Apr 1 at 13:16
add a comment |
$begingroup$
For clarity, are you finding $$int sqrt[3]x(x-1) dx$$ By the way, here's a reference for MathJax you might find useful, which is preferred for writing up and rendering your math text on this site.
$endgroup$
– Eevee Trainer
Apr 1 at 6:22
$begingroup$
Have you tried $x=sec^2 t$?
$endgroup$
– J.G.
Apr 1 at 6:35
$begingroup$
I fixed the equation in the question!
$endgroup$
– ssgg1
Apr 1 at 13:16
$begingroup$
For clarity, are you finding $$int sqrt[3]x(x-1) dx$$ By the way, here's a reference for MathJax you might find useful, which is preferred for writing up and rendering your math text on this site.
$endgroup$
– Eevee Trainer
Apr 1 at 6:22
$begingroup$
For clarity, are you finding $$int sqrt[3]x(x-1) dx$$ By the way, here's a reference for MathJax you might find useful, which is preferred for writing up and rendering your math text on this site.
$endgroup$
– Eevee Trainer
Apr 1 at 6:22
$begingroup$
Have you tried $x=sec^2 t$?
$endgroup$
– J.G.
Apr 1 at 6:35
$begingroup$
Have you tried $x=sec^2 t$?
$endgroup$
– J.G.
Apr 1 at 6:35
$begingroup$
I fixed the equation in the question!
$endgroup$
– ssgg1
Apr 1 at 13:16
$begingroup$
I fixed the equation in the question!
$endgroup$
– ssgg1
Apr 1 at 13:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can simply recall that $sqrt [3] x = x^1/3$, distribute it to both terms in $(x-1)$, and use the power rule.
$$int sqrt[3]x (x-1)dx = int x^1/3(x-1)dx = int (x^4/3 - x^1/3)dx = frac 3 7 x^7/3 - frac 3 4 x^4/3 + C$$
answered Apr 1 at 16:01
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can simply recall that $sqrt [3] x = x^1/3$, distribute it to both terms in $(x-1)$, and use the power rule.
$$int sqrt[3]x (x-1)dx = int x^1/3(x-1)dx = int (x^4/3 - x^1/3)dx = frac 3 7 x^7/3 - frac 3 4 x^4/3 + C$$
answered Apr 1 at 16:01
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
add a comment |
$begingroup$
You can simply recall that $sqrt [3] x = x^1/3$, distribute it to both terms in $(x-1)$, and use the power rule.
$$int sqrt[3]x (x-1)dx = int x^1/3(x-1)dx = int (x^4/3 - x^1/3)dx = frac 3 7 x^7/3 - frac 3 4 x^4/3 + C$$
answered Apr 1 at 16:01
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
add a comment |
$begingroup$
You can simply recall that $sqrt [3] x = x^1/3$, distribute it to both terms in $(x-1)$, and use the power rule.
$$int sqrt[3]x (x-1)dx = int x^1/3(x-1)dx = int (x^4/3 - x^1/3)dx = frac 3 7 x^7/3 - frac 3 4 x^4/3 + C$$
answered Apr 1 at 16:01
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
You can simply recall that $sqrt [3] x = x^1/3$, distribute it to both terms in $(x-1)$, and use the power rule.
$$int sqrt[3]x (x-1)dx = int x^1/3(x-1)dx = int (x^4/3 - x^1/3)dx = frac 3 7 x^7/3 - frac 3 4 x^4/3 + C$$
answered Apr 1 at 16:01
Eevee TrainerEevee Trainer
10.6k31842
answered Apr 1 at 16:01
Eevee TrainerEevee Trainer
10.6k31842
answered Apr 1 at 16:01
Eevee TrainerEevee Trainer
10.6k31842
answered Apr 1 at 16:01
Eevee TrainerEevee Trainer
10.6k31842
10.6k31842
add a comment |
add a comment |
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$begingroup$
For clarity, are you finding $$int sqrt[3]x(x-1) dx$$ By the way, here's a reference for MathJax you might find useful, which is preferred for writing up and rendering your math text on this site.
$endgroup$
– Eevee Trainer
Apr 1 at 6:22
$begingroup$
Have you tried $x=sec^2 t$?
$endgroup$
– J.G.
Apr 1 at 6:35
$begingroup$
I fixed the equation in the question!
$endgroup$
– ssgg1
Apr 1 at 13:16