Two different solutions of $intfrac{1}{1+x} dx$
$begingroup$
I am having doubt in the calculation of the integral
$$
intfrac{1}{1+x} dx,$$
the solution of which is
$$
log(1+x)+C.$$
I have solved this integration in a different way. First I converted the above integral to
$$
;intfrac{1}{1+(sqrt{x})^2} dx.$$
Then I used the formula as
$$
{intfrac{1}{1+x^2} dx}=tan^{-1}x+C
$$
so by using this formula I got as an answer $tan^{-1}(sqrt{x})+C$ which is different from the solution.
Am I correct about this?
calculus integration
asked Dec 26 '18 at 18:47
sanketsanket
32
$endgroup$
add a comment |
$begingroup$
I am having doubt in the calculation of the integral
$$
intfrac{1}{1+x} dx,$$
the solution of which is
$$
log(1+x)+C.$$
I have solved this integration in a different way. First I converted the above integral to
$$
;intfrac{1}{1+(sqrt{x})^2} dx.$$
Then I used the formula as
$$
{intfrac{1}{1+x^2} dx}=tan^{-1}x+C
$$
so by using this formula I got as an answer $tan^{-1}(sqrt{x})+C$ which is different from the solution.
Am I correct about this?
calculus integration
asked Dec 26 '18 at 18:47
sanketsanket
32
$endgroup$
1
$begingroup$
Differentiating $tan^{-1}sqrt x$ gives $frac1{2sqrt x(1+x)}$.
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 18:48
$begingroup$
The correct answer is $ln(|1+x|)+C$.
$endgroup$
– hamam_Abdallah
Dec 26 '18 at 20:02
add a comment |
$begingroup$
I am having doubt in the calculation of the integral
$$
intfrac{1}{1+x} dx,$$
the solution of which is
$$
log(1+x)+C.$$
I have solved this integration in a different way. First I converted the above integral to
$$
;intfrac{1}{1+(sqrt{x})^2} dx.$$
Then I used the formula as
$$
{intfrac{1}{1+x^2} dx}=tan^{-1}x+C
$$
so by using this formula I got as an answer $tan^{-1}(sqrt{x})+C$ which is different from the solution.
Am I correct about this?
calculus integration
asked Dec 26 '18 at 18:47
sanketsanket
32
$endgroup$
I am having doubt in the calculation of the integral
$$
intfrac{1}{1+x} dx,$$
the solution of which is
$$
log(1+x)+C.$$
I have solved this integration in a different way. First I converted the above integral to
$$
;intfrac{1}{1+(sqrt{x})^2} dx.$$
Then I used the formula as
$$
{intfrac{1}{1+x^2} dx}=tan^{-1}x+C
$$
so by using this formula I got as an answer $tan^{-1}(sqrt{x})+C$ which is different from the solution.
Am I correct about this?
calculus integration
calculus integration
asked Dec 26 '18 at 18:47
sanketsanket
32
asked Dec 26 '18 at 18:47
sanketsanket
32
edited Dec 28 '18 at 3:36
user587192
asked Dec 26 '18 at 18:47
sanketsanket
32
asked Dec 26 '18 at 18:47
sanketsanket
32
asked Dec 26 '18 at 18:47
sanketsanket
32
32
1
$begingroup$
Differentiating $tan^{-1}sqrt x$ gives $frac1{2sqrt x(1+x)}$.
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 18:48
$begingroup$
The correct answer is $ln(|1+x|)+C$.
$endgroup$
– hamam_Abdallah
Dec 26 '18 at 20:02
add a comment |
1
$begingroup$
Differentiating $tan^{-1}sqrt x$ gives $frac1{2sqrt x(1+x)}$.
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 18:48
$begingroup$
The correct answer is $ln(|1+x|)+C$.
$endgroup$
– hamam_Abdallah
Dec 26 '18 at 20:02
1
1
$begingroup$
Differentiating $tan^{-1}sqrt x$ gives $frac1{2sqrt x(1+x)}$.
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 18:48
$begingroup$
Differentiating $tan^{-1}sqrt x$ gives $frac1{2sqrt x(1+x)}$.
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 18:48
$begingroup$
The correct answer is $ln(|1+x|)+C$.
$endgroup$
– hamam_Abdallah
Dec 26 '18 at 20:02
$begingroup$
The correct answer is $ln(|1+x|)+C$.
$endgroup$
– hamam_Abdallah
Dec 26 '18 at 20:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The substitution $x=t^2$ (which, by the way, can be done only for $xge0$) brings the integral in the form
$$
intfrac{1}{1+t^2}cdot 2t,dt=log(1+t^2)+c=log(1+x)+c
$$
Recall that integration by substitution is an application of the chain rule for derivatives. The $dx$ is a reminder for you to apply it (more than that, actually).
Otherwise you'd get, similarly to your manipulation,
$$
int x,dx=int (sqrt{x})^2,dx=frac{(sqrt{x})^3}{3}+c
$$
which is clearly absurd. Or
$$
intfrac{1}{sqrt{1-x}},dx=intfrac{1}{sqrt{1-(sqrt{x})^2}},dx=arcsinsqrt{x}+c
$$
whereas the correct antiderivative is $-2sqrt{1-x}+c$.
answered Dec 26 '18 at 19:16
egregegreg
182k1485204
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add a comment |
$begingroup$
No. You forgot to change $rm{d}x$ into $rm{d}(sqrt x)$.
answered Dec 26 '18 at 18:49
Lucas HenriqueLucas Henrique
1,026414
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
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votes
$begingroup$
The substitution $x=t^2$ (which, by the way, can be done only for $xge0$) brings the integral in the form
$$
intfrac{1}{1+t^2}cdot 2t,dt=log(1+t^2)+c=log(1+x)+c
$$
Recall that integration by substitution is an application of the chain rule for derivatives. The $dx$ is a reminder for you to apply it (more than that, actually).
Otherwise you'd get, similarly to your manipulation,
$$
int x,dx=int (sqrt{x})^2,dx=frac{(sqrt{x})^3}{3}+c
$$
which is clearly absurd. Or
$$
intfrac{1}{sqrt{1-x}},dx=intfrac{1}{sqrt{1-(sqrt{x})^2}},dx=arcsinsqrt{x}+c
$$
whereas the correct antiderivative is $-2sqrt{1-x}+c$.
answered Dec 26 '18 at 19:16
egregegreg
182k1485204
$endgroup$
add a comment |
$begingroup$
The substitution $x=t^2$ (which, by the way, can be done only for $xge0$) brings the integral in the form
$$
intfrac{1}{1+t^2}cdot 2t,dt=log(1+t^2)+c=log(1+x)+c
$$
Recall that integration by substitution is an application of the chain rule for derivatives. The $dx$ is a reminder for you to apply it (more than that, actually).
Otherwise you'd get, similarly to your manipulation,
$$
int x,dx=int (sqrt{x})^2,dx=frac{(sqrt{x})^3}{3}+c
$$
which is clearly absurd. Or
$$
intfrac{1}{sqrt{1-x}},dx=intfrac{1}{sqrt{1-(sqrt{x})^2}},dx=arcsinsqrt{x}+c
$$
whereas the correct antiderivative is $-2sqrt{1-x}+c$.
answered Dec 26 '18 at 19:16
egregegreg
182k1485204
$endgroup$
add a comment |
$begingroup$
The substitution $x=t^2$ (which, by the way, can be done only for $xge0$) brings the integral in the form
$$
intfrac{1}{1+t^2}cdot 2t,dt=log(1+t^2)+c=log(1+x)+c
$$
Recall that integration by substitution is an application of the chain rule for derivatives. The $dx$ is a reminder for you to apply it (more than that, actually).
Otherwise you'd get, similarly to your manipulation,
$$
int x,dx=int (sqrt{x})^2,dx=frac{(sqrt{x})^3}{3}+c
$$
which is clearly absurd. Or
$$
intfrac{1}{sqrt{1-x}},dx=intfrac{1}{sqrt{1-(sqrt{x})^2}},dx=arcsinsqrt{x}+c
$$
whereas the correct antiderivative is $-2sqrt{1-x}+c$.
answered Dec 26 '18 at 19:16
egregegreg
182k1485204
$endgroup$
The substitution $x=t^2$ (which, by the way, can be done only for $xge0$) brings the integral in the form
$$
intfrac{1}{1+t^2}cdot 2t,dt=log(1+t^2)+c=log(1+x)+c
$$
Recall that integration by substitution is an application of the chain rule for derivatives. The $dx$ is a reminder for you to apply it (more than that, actually).
Otherwise you'd get, similarly to your manipulation,
$$
int x,dx=int (sqrt{x})^2,dx=frac{(sqrt{x})^3}{3}+c
$$
which is clearly absurd. Or
$$
intfrac{1}{sqrt{1-x}},dx=intfrac{1}{sqrt{1-(sqrt{x})^2}},dx=arcsinsqrt{x}+c
$$
whereas the correct antiderivative is $-2sqrt{1-x}+c$.
answered Dec 26 '18 at 19:16
egregegreg
182k1485204
answered Dec 26 '18 at 19:16
egregegreg
182k1485204
answered Dec 26 '18 at 19:16
egregegreg
182k1485204
answered Dec 26 '18 at 19:16
egregegreg
182k1485204
182k1485204
add a comment |
add a comment |
$begingroup$
No. You forgot to change $rm{d}x$ into $rm{d}(sqrt x)$.
answered Dec 26 '18 at 18:49
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
add a comment |
$begingroup$
No. You forgot to change $rm{d}x$ into $rm{d}(sqrt x)$.
answered Dec 26 '18 at 18:49
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
add a comment |
$begingroup$
No. You forgot to change $rm{d}x$ into $rm{d}(sqrt x)$.
answered Dec 26 '18 at 18:49
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
No. You forgot to change $rm{d}x$ into $rm{d}(sqrt x)$.
answered Dec 26 '18 at 18:49
Lucas HenriqueLucas Henrique
1,026414
answered Dec 26 '18 at 18:49
Lucas HenriqueLucas Henrique
1,026414
answered Dec 26 '18 at 18:49
Lucas HenriqueLucas Henrique
1,026414
answered Dec 26 '18 at 18:49
Lucas HenriqueLucas Henrique
1,026414
1,026414
add a comment |
add a comment |
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1
$begingroup$
Differentiating $tan^{-1}sqrt x$ gives $frac1{2sqrt x(1+x)}$.
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 18:48
$begingroup$
The correct answer is $ln(|1+x|)+C$.
$endgroup$
– hamam_Abdallah
Dec 26 '18 at 20:02