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Integral with two distinct roots in them, such as: $int fracsqrtxx^2(sqrtx+1+sqrtx)dx$

Indefinite integral $intfracdxx^2+2$Simple integral with a logarithmic function: $intfracln xsqrt1-x,mathrm dx$Integral $int fracsqrt16-x^2x mathrmdx$Problem with Indefinite Integral $intfrac cos^4xsin^3x} dx$Problem with Indefinite Integral $int frac cos^5x 16(cos^4x+sin^4x)dx$How do I solve the integral $intfrac{1-sqrt2x+31+sqrt2x+3dx$ with help substitution?Solving trigonometric indefinite integral $int fracdxsqrttan x $Solving Indefinite Integral $intfracx^2sqrt1-xdx$Computing a double integral: $int _0^frac14int _sqrtx^frac12:fraccosleft(pi yright)y^2~mathrm dy~mathrm dx$Does my book lie about $int fracarccos(fracx2)sqrt4-x^2dx$?

0

$begingroup$

I'm getting familiar with basic indefinite integrals and these are the hardest ones I've met so far:

1. $int fracsqrtxx^2(sqrtx+1+sqrtx)dx$




2. $int fracsqrt[3]x+2-sqrt[3]xx^2(sqrt[3]x+2+sqrt[3]x)$

Any hints? Please note that the course I am taking does not anticipate usage of hyperbolic functions. I am not familiar with them.

The first integral I attempted:

$int fracsqrtxx^2(sqrtx+1+sqrtx)dx = int fracsqrtx(sqrtx+1-sqrtx)x^2(sqrtx+1+sqrtx)(sqrtx+1-sqrtx)dx = int fracsqrtxsqrtx+1-xx^2dx$

Now I can split it into two integrals. Problem is with:

$int fracsqrtxsqrtx+1x^2dx$

and the major problem is that I don't know how to solve integrals that have some distinct roots with different values inside those roots. Second task's integral seems even harder.

If speaking of "different values under roots", I am only familiar with how to solve such integrals:

$int frac(sqrtfracx+2x-1-1)^23(sqrtfracx+2x-1+2)dx$

because there's simple algorithm that I can follow to solve it.

Hints, tips, advices appreciated. Thanks.

EDIT: $int fracsqrtxsqrtx+1x^2dx = int fracsqrtx^2+xx^2dx = int fracx^2 + x x^2sqrtx^2+xdx = int fracx+1xsqrtx^2+xdx$

And now with Euler's substitution should work?

calculus integration indefinite-integrals

share|cite|improve this question

edited Mar 29 at 17:10

weno

asked Mar 29 at 17:00

wenoweno

39611

$endgroup$

add a comment | 

0

$begingroup$

I'm getting familiar with basic indefinite integrals and these are the hardest ones I've met so far:

1. $int fracsqrtxx^2(sqrtx+1+sqrtx)dx$




2. $int fracsqrt[3]x+2-sqrt[3]xx^2(sqrt[3]x+2+sqrt[3]x)$

Any hints? Please note that the course I am taking does not anticipate usage of hyperbolic functions. I am not familiar with them.

The first integral I attempted:

$int fracsqrtxx^2(sqrtx+1+sqrtx)dx = int fracsqrtx(sqrtx+1-sqrtx)x^2(sqrtx+1+sqrtx)(sqrtx+1-sqrtx)dx = int fracsqrtxsqrtx+1-xx^2dx$

Now I can split it into two integrals. Problem is with:

$int fracsqrtxsqrtx+1x^2dx$

and the major problem is that I don't know how to solve integrals that have some distinct roots with different values inside those roots. Second task's integral seems even harder.

If speaking of "different values under roots", I am only familiar with how to solve such integrals:

$int frac(sqrtfracx+2x-1-1)^23(sqrtfracx+2x-1+2)dx$

because there's simple algorithm that I can follow to solve it.

Hints, tips, advices appreciated. Thanks.

EDIT: $int fracsqrtxsqrtx+1x^2dx = int fracsqrtx^2+xx^2dx = int fracx^2 + x x^2sqrtx^2+xdx = int fracx+1xsqrtx^2+xdx$

And now with Euler's substitution should work?

calculus integration indefinite-integrals

share|cite|improve this question

edited Mar 29 at 17:10

weno

asked Mar 29 at 17:00

wenoweno

39611

$endgroup$

add a comment | 

$begingroup$

I'm getting familiar with basic indefinite integrals and these are the hardest ones I've met so far:

1. $int fracsqrtxx^2(sqrtx+1+sqrtx)dx$




2. $int fracsqrt[3]x+2-sqrt[3]xx^2(sqrt[3]x+2+sqrt[3]x)$

Any hints? Please note that the course I am taking does not anticipate usage of hyperbolic functions. I am not familiar with them.

The first integral I attempted:

$int fracsqrtxx^2(sqrtx+1+sqrtx)dx = int fracsqrtx(sqrtx+1-sqrtx)x^2(sqrtx+1+sqrtx)(sqrtx+1-sqrtx)dx = int fracsqrtxsqrtx+1-xx^2dx$

Now I can split it into two integrals. Problem is with:

$int fracsqrtxsqrtx+1x^2dx$

and the major problem is that I don't know how to solve integrals that have some distinct roots with different values inside those roots. Second task's integral seems even harder.

If speaking of "different values under roots", I am only familiar with how to solve such integrals:

$int frac(sqrtfracx+2x-1-1)^23(sqrtfracx+2x-1+2)dx$

because there's simple algorithm that I can follow to solve it.

Hints, tips, advices appreciated. Thanks.

EDIT: $int fracsqrtxsqrtx+1x^2dx = int fracsqrtx^2+xx^2dx = int fracx^2 + x x^2sqrtx^2+xdx = int fracx+1xsqrtx^2+xdx$

And now with Euler's substitution should work?

calculus integration indefinite-integrals

share|cite|improve this question

edited Mar 29 at 17:10

weno

asked Mar 29 at 17:00

wenoweno

39611

$endgroup$

share|cite|improve this question

edited Mar 29 at 17:10

weno

asked Mar 29 at 17:00

wenoweno

39611

share|cite|improve this question

edited Mar 29 at 17:10

weno

asked Mar 29 at 17:00

wenoweno

39611

asked Mar 29 at 17:00

wenoweno

39611

asked Mar 29 at 17:00

wenoweno

39611

1 Answer
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Hint: For the integral $$int fracsqrtx^2+xx^2dx$$ substitute $$sqrtx^2+x=x+t$$ it is the Eulerian substitution.
Then we get by squaring $$x=fract^21-2t$$ and $$dx=-2,frac t left( -1+t right) left( -1+2,t right) ^2dt$$

share|cite|improve this answer

edited Mar 29 at 17:46

answered Mar 29 at 17:19

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.6k42867

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1

$begingroup$

Hint: For the integral $$int fracsqrtx^2+xx^2dx$$ substitute $$sqrtx^2+x=x+t$$ it is the Eulerian substitution.
Then we get by squaring $$x=fract^21-2t$$ and $$dx=-2,frac t left( -1+t right) left( -1+2,t right) ^2dt$$

share|cite|improve this answer

edited Mar 29 at 17:46

answered Mar 29 at 17:19

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.6k42867

$endgroup$

add a comment | 

1

$begingroup$

Hint: For the integral $$int fracsqrtx^2+xx^2dx$$ substitute $$sqrtx^2+x=x+t$$ it is the Eulerian substitution.
Then we get by squaring $$x=fract^21-2t$$ and $$dx=-2,frac t left( -1+t right) left( -1+2,t right) ^2dt$$

share|cite|improve this answer

edited Mar 29 at 17:46

answered Mar 29 at 17:19

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.6k42867

$endgroup$

add a comment | 

$begingroup$

Hint: For the integral $$int fracsqrtx^2+xx^2dx$$ substitute $$sqrtx^2+x=x+t$$ it is the Eulerian substitution.
Then we get by squaring $$x=fract^21-2t$$ and $$dx=-2,frac t left( -1+t right) left( -1+2,t right) ^2dt$$

share|cite|improve this answer

edited Mar 29 at 17:46

answered Mar 29 at 17:19

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.6k42867

$endgroup$

share|cite|improve this answer

edited Mar 29 at 17:46

answered Mar 29 at 17:19

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.6k42867

answered Mar 29 at 17:19

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.6k42867

answered Mar 29 at 17:19

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.6k42867