Irrational integral $int frac1sqrtx sqrtfracsqrtx-2sqrtx+2dx$ The Next CEO of Stack OverflowIntegral of $int^1_0fracdxsqrtx+3-1$Compute the integral $intfracdx(x^2-x+1)sqrtx^2+x+1$Evaluation of the integral $int sqrtt^4-t^2 + 1,dt$Evaluating trigonometric integrals of the form $int fracC ; d theta sin theta sqrtsin^2 theta - C $Choosing a substitution to evaluate $int fracx+3sqrtx+2dx$Solve integral $int fracsqrtx+1+2(x+1)^2 - sqrtx+1dx$Integral $int sqrtfracx2-xdx$Indefinite integral $int frac2^x2^2x-4dx$Calculate $int frac1x^2+x+1 , dx$Integral $intfrac1sqrt2x^2+x+1dx$
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Irrational integral $int frac1sqrtx sqrtfracsqrtx-2sqrtx+2dx$
The Next CEO of Stack OverflowIntegral of $int^1_0fracdxsqrtx+3-1$Compute the integral $intfracdx(x^2-x+1)sqrtx^2+x+1$Evaluation of the integral $int sqrtt^4-t^2 + 1,dt$Evaluating trigonometric integrals of the form $int fracC ; d theta sin theta sqrtsin^2 theta - C $Choosing a substitution to evaluate $int fracx+3sqrtx+2dx$Solve integral $int fracsqrtx+1+2(x+1)^2 - sqrtx+1dx$Integral $int sqrtfracx2-xdx$Indefinite integral $int frac2^x2^2x-4dx$Calculate $int frac1x^2+x+1 , dx$Integral $intfrac1sqrt2x^2+x+1dx$
$begingroup$
Would anyone be able to verify if this integral is calculated correctly?
$$int frac1sqrtx sqrtfracsqrtx-2sqrtx+2dx$$
My attempt:
substitute:$left(t = sqrtx, t^2 = x, 2tdt=dx right)$
$$
beginsplit
int frac1t sqrtfract-2t+2,2t,dt
&= 2intsqrtfract-2t+2dt
= 2int fracsqrtt-2sqrtt+2dt \
&= 2int fracsqrt(t-2)(t-2)sqrt(t+2)(t-2)dt \
&= 2int fract-2sqrtt^2-4dt \
&= intfrac2t-4sqrtt^2-4dt \
&= int frac2tsqrtt^2-4dt - 4intfracdtsqrtt^2-4 \
&= sqrtt^2-4 - 4ln + C
endsplit
$$
Substitute back $t = sqrtx$:
result: $Longrightarrow sqrtx-4 - 4ln + C$
calculus integration indefinite-integrals
edited Mar 27 at 18:36
Maria Mazur
49k1360122
asked Mar 27 at 18:24
wenoweno
36011
$endgroup$
add a comment |
$begingroup$
Would anyone be able to verify if this integral is calculated correctly?
$$int frac1sqrtx sqrtfracsqrtx-2sqrtx+2dx$$
My attempt:
substitute:$left(t = sqrtx, t^2 = x, 2tdt=dx right)$
$$
beginsplit
int frac1t sqrtfract-2t+2,2t,dt
&= 2intsqrtfract-2t+2dt
= 2int fracsqrtt-2sqrtt+2dt \
&= 2int fracsqrt(t-2)(t-2)sqrt(t+2)(t-2)dt \
&= 2int fract-2sqrtt^2-4dt \
&= intfrac2t-4sqrtt^2-4dt \
&= int frac2tsqrtt^2-4dt - 4intfracdtsqrtt^2-4 \
&= sqrtt^2-4 - 4ln + C
endsplit
$$
Substitute back $t = sqrtx$:
result: $Longrightarrow sqrtx-4 - 4ln + C$
calculus integration indefinite-integrals
edited Mar 27 at 18:36
Maria Mazur
49k1360122
asked Mar 27 at 18:24
wenoweno
36011
$endgroup$
1
$begingroup$
You are missing a factor of $2$ on the $sqrtt^2-4$ term.
$endgroup$
– Peter Foreman
Mar 27 at 18:31
$begingroup$
That's right, thanks.
$endgroup$
– weno
Mar 27 at 18:34
add a comment |
$begingroup$
Would anyone be able to verify if this integral is calculated correctly?
$$int frac1sqrtx sqrtfracsqrtx-2sqrtx+2dx$$
My attempt:
substitute:$left(t = sqrtx, t^2 = x, 2tdt=dx right)$
$$
beginsplit
int frac1t sqrtfract-2t+2,2t,dt
&= 2intsqrtfract-2t+2dt
= 2int fracsqrtt-2sqrtt+2dt \
&= 2int fracsqrt(t-2)(t-2)sqrt(t+2)(t-2)dt \
&= 2int fract-2sqrtt^2-4dt \
&= intfrac2t-4sqrtt^2-4dt \
&= int frac2tsqrtt^2-4dt - 4intfracdtsqrtt^2-4 \
&= sqrtt^2-4 - 4ln + C
endsplit
$$
Substitute back $t = sqrtx$:
result: $Longrightarrow sqrtx-4 - 4ln + C$
calculus integration indefinite-integrals
edited Mar 27 at 18:36
Maria Mazur
49k1360122
asked Mar 27 at 18:24
wenoweno
36011
$endgroup$
Would anyone be able to verify if this integral is calculated correctly?
$$int frac1sqrtx sqrtfracsqrtx-2sqrtx+2dx$$
My attempt:
substitute:$left(t = sqrtx, t^2 = x, 2tdt=dx right)$
$$
beginsplit
int frac1t sqrtfract-2t+2,2t,dt
&= 2intsqrtfract-2t+2dt
= 2int fracsqrtt-2sqrtt+2dt \
&= 2int fracsqrt(t-2)(t-2)sqrt(t+2)(t-2)dt \
&= 2int fract-2sqrtt^2-4dt \
&= intfrac2t-4sqrtt^2-4dt \
&= int frac2tsqrtt^2-4dt - 4intfracdtsqrtt^2-4 \
&= sqrtt^2-4 - 4ln + C
endsplit
$$
Substitute back $t = sqrtx$:
result: $Longrightarrow sqrtx-4 - 4ln + C$
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Mar 27 at 18:36
Maria Mazur
49k1360122
asked Mar 27 at 18:24
wenoweno
36011
edited Mar 27 at 18:36
Maria Mazur
49k1360122
asked Mar 27 at 18:24
wenoweno
36011
edited Mar 27 at 18:36
Maria Mazur
49k1360122
edited Mar 27 at 18:36
Maria Mazur
49k1360122
edited Mar 27 at 18:36
Maria Mazur
49k1360122
49k1360122
asked Mar 27 at 18:24
wenoweno
36011
asked Mar 27 at 18:24
wenoweno
36011
asked Mar 27 at 18:24
wenoweno
36011
36011
1
$begingroup$
You are missing a factor of $2$ on the $sqrtt^2-4$ term.
$endgroup$
– Peter Foreman
Mar 27 at 18:31
$begingroup$
That's right, thanks.
$endgroup$
– weno
Mar 27 at 18:34
add a comment |
1
$begingroup$
You are missing a factor of $2$ on the $sqrtt^2-4$ term.
$endgroup$
– Peter Foreman
Mar 27 at 18:31
$begingroup$
That's right, thanks.
$endgroup$
– weno
Mar 27 at 18:34
1
1
$begingroup$
You are missing a factor of $2$ on the $sqrtt^2-4$ term.
$endgroup$
– Peter Foreman
Mar 27 at 18:31
$begingroup$
You are missing a factor of $2$ on the $sqrtt^2-4$ term.
$endgroup$
– Peter Foreman
Mar 27 at 18:31
$begingroup$
That's right, thanks.
$endgroup$
– weno
Mar 27 at 18:34
$begingroup$
That's right, thanks.
$endgroup$
– weno
Mar 27 at 18:34
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$...= colorred2sqrtt^2-4 - 4ln + C$$
Anyway, you could take a derivate and check it you self.
answered Mar 27 at 18:29
Maria MazurMaria Mazur
49k1360122
$endgroup$
1
$begingroup$
Thanks. I will accept in 4 minutes.
$endgroup$
– weno
Mar 27 at 18:34
add a comment |
$begingroup$
Any time you want to check your answer for an integral go to https://www.integral-calculator.com/
answered Mar 27 at 18:31
Rithik KapoorRithik Kapoor
31010
$endgroup$
add a comment |
$begingroup$
I got $2sqrtx-4-8operatornamearcsinhleft(fracsqrtsqrtx-22right)+C$
answered Mar 27 at 18:30
El EctricEl Ectric
14511
$endgroup$
3
$begingroup$
Note that $operatornamearcsinhphi=lnleft(phi+sqrtphi^2+1right)$.
$endgroup$
– J.G.
Mar 27 at 18:32
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$...= colorred2sqrtt^2-4 - 4ln + C$$
Anyway, you could take a derivate and check it you self.
answered Mar 27 at 18:29
Maria MazurMaria Mazur
49k1360122
$endgroup$
1
$begingroup$
Thanks. I will accept in 4 minutes.
$endgroup$
– weno
Mar 27 at 18:34
add a comment |
$begingroup$
$$...= colorred2sqrtt^2-4 - 4ln + C$$
Anyway, you could take a derivate and check it you self.
answered Mar 27 at 18:29
Maria MazurMaria Mazur
49k1360122
$endgroup$
1
$begingroup$
Thanks. I will accept in 4 minutes.
$endgroup$
– weno
Mar 27 at 18:34
add a comment |
$begingroup$
$$...= colorred2sqrtt^2-4 - 4ln + C$$
Anyway, you could take a derivate and check it you self.
answered Mar 27 at 18:29
Maria MazurMaria Mazur
49k1360122
$endgroup$
$$...= colorred2sqrtt^2-4 - 4ln + C$$
Anyway, you could take a derivate and check it you self.
answered Mar 27 at 18:29
Maria MazurMaria Mazur
49k1360122
answered Mar 27 at 18:29
Maria MazurMaria Mazur
49k1360122
answered Mar 27 at 18:29
Maria MazurMaria Mazur
49k1360122
answered Mar 27 at 18:29
Maria MazurMaria Mazur
49k1360122
49k1360122
1
$begingroup$
Thanks. I will accept in 4 minutes.
$endgroup$
– weno
Mar 27 at 18:34
add a comment |
1
$begingroup$
Thanks. I will accept in 4 minutes.
$endgroup$
– weno
Mar 27 at 18:34
1
1
$begingroup$
Thanks. I will accept in 4 minutes.
$endgroup$
– weno
Mar 27 at 18:34
$begingroup$
Thanks. I will accept in 4 minutes.
$endgroup$
– weno
Mar 27 at 18:34
add a comment |
$begingroup$
Any time you want to check your answer for an integral go to https://www.integral-calculator.com/
answered Mar 27 at 18:31
Rithik KapoorRithik Kapoor
31010
$endgroup$
add a comment |
$begingroup$
Any time you want to check your answer for an integral go to https://www.integral-calculator.com/
answered Mar 27 at 18:31
Rithik KapoorRithik Kapoor
31010
$endgroup$
add a comment |
$begingroup$
Any time you want to check your answer for an integral go to https://www.integral-calculator.com/
answered Mar 27 at 18:31
Rithik KapoorRithik Kapoor
31010
$endgroup$
Any time you want to check your answer for an integral go to https://www.integral-calculator.com/
answered Mar 27 at 18:31
Rithik KapoorRithik Kapoor
31010
answered Mar 27 at 18:31
Rithik KapoorRithik Kapoor
31010
answered Mar 27 at 18:31
Rithik KapoorRithik Kapoor
31010
answered Mar 27 at 18:31
Rithik KapoorRithik Kapoor
31010
31010
add a comment |
add a comment |
$begingroup$
I got $2sqrtx-4-8operatornamearcsinhleft(fracsqrtsqrtx-22right)+C$
answered Mar 27 at 18:30
El EctricEl Ectric
14511
$endgroup$
3
$begingroup$
Note that $operatornamearcsinhphi=lnleft(phi+sqrtphi^2+1right)$.
$endgroup$
– J.G.
Mar 27 at 18:32
add a comment |
$begingroup$
I got $2sqrtx-4-8operatornamearcsinhleft(fracsqrtsqrtx-22right)+C$
answered Mar 27 at 18:30
El EctricEl Ectric
14511
$endgroup$
3
$begingroup$
Note that $operatornamearcsinhphi=lnleft(phi+sqrtphi^2+1right)$.
$endgroup$
– J.G.
Mar 27 at 18:32
add a comment |
$begingroup$
I got $2sqrtx-4-8operatornamearcsinhleft(fracsqrtsqrtx-22right)+C$
answered Mar 27 at 18:30
El EctricEl Ectric
14511
$endgroup$
I got $2sqrtx-4-8operatornamearcsinhleft(fracsqrtsqrtx-22right)+C$
answered Mar 27 at 18:30
El EctricEl Ectric
14511
edited Mar 27 at 18:32
answered Mar 27 at 18:30
El EctricEl Ectric
14511
answered Mar 27 at 18:30
El EctricEl Ectric
14511
answered Mar 27 at 18:30
El EctricEl Ectric
14511
14511
3
$begingroup$
Note that $operatornamearcsinhphi=lnleft(phi+sqrtphi^2+1right)$.
$endgroup$
– J.G.
Mar 27 at 18:32
add a comment |
3
$begingroup$
Note that $operatornamearcsinhphi=lnleft(phi+sqrtphi^2+1right)$.
$endgroup$
– J.G.
Mar 27 at 18:32
3
3
$begingroup$
Note that $operatornamearcsinhphi=lnleft(phi+sqrtphi^2+1right)$.
$endgroup$
– J.G.
Mar 27 at 18:32
$begingroup$
Note that $operatornamearcsinhphi=lnleft(phi+sqrtphi^2+1right)$.
$endgroup$
– J.G.
Mar 27 at 18:32
add a comment |
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1
$begingroup$
You are missing a factor of $2$ on the $sqrtt^2-4$ term.
$endgroup$
– Peter Foreman
Mar 27 at 18:31
$begingroup$
That's right, thanks.
$endgroup$
– weno
Mar 27 at 18:34