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int_(Gamma)(e^(2))/(z*(2^(2)+1))dz

$\int_{\Gamma} \frac{e^{2}}{z \cdot\left(2^{2}+1\right)} d z$

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Find indefinite integrals using the substitution

Full solution

Q.

\int_{\Gamma} \frac{e^{2}}{z \cdot\left(2^{2}+1\right)} d z

Simplify the integral: Let's start by simplifying the integral a bit, we have: $\newline$ $\int\frac{e^{2}}{z(z^{2}+1)}dz$ $\newline$ Notice that $e^{2}$ is a constant and can be factored out of the integral. $\newline$ So, we get: $\newline$ $e^{2} \cdot \int\frac{1}{z(z^{2}+1)}dz$
Partial fraction decomposition: Now, let's do a partial fraction decomposition on $\frac{1}{z(z^{2}+1)}$ . We want to express $\frac{1}{z(z^{2}+1)}$ as $\frac{A}{z} + \frac{Bz+C}{z^{2}+1}$ . To find $A$ , $B$ , and $C$ , we multiply both sides by $z(z^{2}+1)$ and compare coefficients. $1 = A(z^{2}+1) + (Bz+C)z$
Find coefficients: Setting $z=0$ to find $A$ gives us: $\newline$ $1 = A*(0+1)$ $\newline$ So, $A = 1$ .
Integrate term by term: Now we need to find $B$ and $C$ . We can plug in different values for $z$ or compare coefficients. Let's compare coefficients for the $z^2$ , $z$ , and constant terms. For $z^2$ terms: $0 = A + B$ For $z$ terms: $0 = C$ For constant terms: $1 = A$ We already know $C$ $0$ , so $C$ $1$ and $C$ $2$ .
Substitute and integrate: Now we have our partial fraction decomposition: $\newline$ $\frac{1}{z(z^{2}+1)} = \frac{1}{z} - \frac{z}{z^{2}+1}$ $\newline$ So the integral becomes: $\newline$ $e^{2} \cdot \int(\frac{1}{z} - \frac{z}{z^{2}+1})dz$
Combine logarithms: Let's integrate term by term. $\newline$ The integral of $\frac{1}{z} \, dz$ is $\ln|z|$ . $\newline$ The integral of $-\frac{z}{z^{2}+1} \, dz$ is a bit trickier, but we can use a substitution. $\newline$ Let $u = z^2 + 1$ , then $du = 2z \, dz$ .
Final simplification: We substitute and get: $\newline$ $\int -\frac{z}{z^{2}+1} dz = -\frac{1}{2} \int \frac{1}{u} du$ $\newline$ Which integrates to $-\frac{1}{2} \ln|u|$ .
Final simplification: We substitute and get: $\newline$ $\int -\frac{z}{z^{2}+1} dz = -\frac{1}{2} \int \frac{1}{u} du$ $\newline$ Which integrates to $-\frac{1}{2} \ln|u|$ .Now we substitute back for u to get the integral in terms of z. $\newline$ $-\frac{1}{2} \ln|u|$ becomes $-\frac{1}{2} \ln|z^{2} + 1|$ .
Final simplification: We substitute and get: $\newline$ $\int -\frac{z}{z^{2}+1} dz = -\frac{1}{2} \int \frac{1}{u} du$ $\newline$ Which integrates to $-\frac{1}{2} \ln|u|$ .Now we substitute back for u to get the integral in terms of z. $\newline$ $-\frac{1}{2} \ln|u|$ becomes $-\frac{1}{2} \ln|z^2 + 1|$ .Putting it all together, we have: $\newline$ $e^2 \cdot (\ln|z| - \frac{1}{2} \ln|z^2 + 1|) + C$
Final simplification: We substitute and get: $\newline$ $\int -\frac{z}{z^{2}+1} dz = -\frac{1}{2} \int \frac{1}{u} du$ $\newline$ Which integrates to $-\frac{1}{2} \ln|u|$ .Now we substitute back for u to get the integral in terms of z. $\newline$ $-\frac{1}{2} \ln|u|$ becomes $-\frac{1}{2} \ln|z^{2} + 1|$ .Putting it all together, we have: $\newline$ $e^{2} \cdot (\ln|z| - \frac{1}{2} \ln|z^{2} + 1|) + C$ Finally, we can simplify a bit by combining the logarithms: $\newline$ $e^{2} \cdot \ln|\frac{z}{(z^{2} + 1)^{\frac{1}{2}}}| + C$

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