What is the integral of $logleft(x^{2} + k^{2}right)$ ?.
$begingroup$
Where $displaystyle k$ is a real number ?. I have tried everything but I stuck when I have to find integral of
$k^{2}/left(x^{2} + k^{2}right)$.
real-analysis analysis
edited Dec 10 '18 at 19:26
Felix Marin
68.1k7108142
asked Dec 10 '18 at 15:49
Panagiotis PanagiotouPanagiotis Panagiotou
1
$endgroup$
add a comment |
$begingroup$
Where $displaystyle k$ is a real number ?. I have tried everything but I stuck when I have to find integral of
$k^{2}/left(x^{2} + k^{2}right)$.
real-analysis analysis
edited Dec 10 '18 at 19:26
Felix Marin
68.1k7108142
asked Dec 10 '18 at 15:49
Panagiotis PanagiotouPanagiotis Panagiotou
1
$endgroup$
2
$begingroup$
You may try $x^2 + k^2 = (x - jk) (x + jk)$ ...
$endgroup$
– Damien
Dec 10 '18 at 16:17
1
$begingroup$
try the substitution $x = ktan theta$
$endgroup$
– Doug M
Dec 10 '18 at 18:10
add a comment |
$begingroup$
Where $displaystyle k$ is a real number ?. I have tried everything but I stuck when I have to find integral of
$k^{2}/left(x^{2} + k^{2}right)$.
real-analysis analysis
edited Dec 10 '18 at 19:26
Felix Marin
68.1k7108142
asked Dec 10 '18 at 15:49
Panagiotis PanagiotouPanagiotis Panagiotou
1
$endgroup$
Where $displaystyle k$ is a real number ?. I have tried everything but I stuck when I have to find integral of
$k^{2}/left(x^{2} + k^{2}right)$.
real-analysis analysis
real-analysis analysis
edited Dec 10 '18 at 19:26
Felix Marin
68.1k7108142
asked Dec 10 '18 at 15:49
Panagiotis PanagiotouPanagiotis Panagiotou
1
edited Dec 10 '18 at 19:26
Felix Marin
68.1k7108142
asked Dec 10 '18 at 15:49
Panagiotis PanagiotouPanagiotis Panagiotou
1
edited Dec 10 '18 at 19:26
Felix Marin
68.1k7108142
edited Dec 10 '18 at 19:26
Felix Marin
68.1k7108142
edited Dec 10 '18 at 19:26
Felix Marin
68.1k7108142
68.1k7108142
asked Dec 10 '18 at 15:49
Panagiotis PanagiotouPanagiotis Panagiotou
1
asked Dec 10 '18 at 15:49
Panagiotis PanagiotouPanagiotis Panagiotou
1
asked Dec 10 '18 at 15:49
Panagiotis PanagiotouPanagiotis Panagiotou
1
1
2
$begingroup$
You may try $x^2 + k^2 = (x - jk) (x + jk)$ ...
$endgroup$
– Damien
Dec 10 '18 at 16:17
1
$begingroup$
try the substitution $x = ktan theta$
$endgroup$
– Doug M
Dec 10 '18 at 18:10
add a comment |
2
$begingroup$
You may try $x^2 + k^2 = (x - jk) (x + jk)$ ...
$endgroup$
– Damien
Dec 10 '18 at 16:17
1
$begingroup$
try the substitution $x = ktan theta$
$endgroup$
– Doug M
Dec 10 '18 at 18:10
2
2
$begingroup$
You may try $x^2 + k^2 = (x - jk) (x + jk)$ ...
$endgroup$
– Damien
Dec 10 '18 at 16:17
$begingroup$
You may try $x^2 + k^2 = (x - jk) (x + jk)$ ...
$endgroup$
– Damien
Dec 10 '18 at 16:17
1
1
$begingroup$
try the substitution $x = ktan theta$
$endgroup$
– Doug M
Dec 10 '18 at 18:10
$begingroup$
try the substitution $x = ktan theta$
$endgroup$
– Doug M
Dec 10 '18 at 18:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Integrate by parts. $u=ln(x^2+k^2)$ and $dv=dx$ to change to the integral
$$int frac{2x^2}{x^2+k^2} ; dx$$
which is do-able by ordinary methods.
answered Dec 10 '18 at 19:35
B. GoddardB. Goddard
19k21440
$endgroup$
add a comment |
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$begingroup$
Integrate by parts. $u=ln(x^2+k^2)$ and $dv=dx$ to change to the integral
$$int frac{2x^2}{x^2+k^2} ; dx$$
which is do-able by ordinary methods.
answered Dec 10 '18 at 19:35
B. GoddardB. Goddard
19k21440
$endgroup$
add a comment |
$begingroup$
Integrate by parts. $u=ln(x^2+k^2)$ and $dv=dx$ to change to the integral
$$int frac{2x^2}{x^2+k^2} ; dx$$
which is do-able by ordinary methods.
answered Dec 10 '18 at 19:35
B. GoddardB. Goddard
19k21440
$endgroup$
add a comment |
$begingroup$
Integrate by parts. $u=ln(x^2+k^2)$ and $dv=dx$ to change to the integral
$$int frac{2x^2}{x^2+k^2} ; dx$$
which is do-able by ordinary methods.
answered Dec 10 '18 at 19:35
B. GoddardB. Goddard
19k21440
$endgroup$
Integrate by parts. $u=ln(x^2+k^2)$ and $dv=dx$ to change to the integral
$$int frac{2x^2}{x^2+k^2} ; dx$$
which is do-able by ordinary methods.
answered Dec 10 '18 at 19:35
B. GoddardB. Goddard
19k21440
answered Dec 10 '18 at 19:35
B. GoddardB. Goddard
19k21440
answered Dec 10 '18 at 19:35
B. GoddardB. Goddard
19k21440
answered Dec 10 '18 at 19:35
B. GoddardB. Goddard
19k21440
19k21440
add a comment |
add a comment |
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2
$begingroup$
You may try $x^2 + k^2 = (x - jk) (x + jk)$ ...
$endgroup$
– Damien
Dec 10 '18 at 16:17
1
$begingroup$
try the substitution $x = ktan theta$
$endgroup$
– Doug M
Dec 10 '18 at 18:10