Question Statement

Evaluate the integral:

$\int \frac{e^{{\tan }^{-1}z}}{1+z^2}dz$

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Background and Explanation

This problem requires the method of substitution (u-substitution). The key insight is recognizing that the derivative of ${\tan }^{-1}z$ is $\frac{1}{1+z^2}$, which appears as a factor in the integrand, allowing us to simplify the integral significantly.

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Solution

We begin by setting up the integral:

$I=\int \frac{e^{{\tan }^{-1}z}}{1+z^2}dz$

To solve this, we use substitution. Let:

$t={\tan }^{-1}z$

Differentiating both sides with respect to $z$:

$\frac{dt}{dz}=\frac{1}{1+z^2}$

Rearranging to express $dz$ in terms of $dt$:

$dt=\frac{dz}{1+z^2}\text{or}dz=(1+z^2)dt$

Now substitute $t={\tan }^{-1}z$ and $dt=\frac{dz}{1+z^2}$ into the original integral:

$I=\int e^{t}dt$

This is a standard exponential integral:

$I=e^t+C$

Finally, substitute back $t={\tan }^{-1}z$ to express the answer in terms of the original variable $z$:

$I=e^{{\tan }^{-1}z}+C$

Therefore:

$\int \frac{e^{{\tan }^{-1}z}}{1+z^2}dz=e^{{\tan }^{-1}z}+C$

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Key Formulas or Methods Used

• Substitution Rule: $\int f(g(z))\cdot g^{\prime }(z)dz=\int f(t)dt$ where $t=g(z)$
• Derivative of Inverse Tangent: $\frac{d}{dz}({\tan }^{-1}z)=\frac{1}{1+z^2}$
• Exponential Integral: $\int e^{x}dx=e^x+C$

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Summary of Steps

1. Identify the substitution: Let $t={\tan }^{-1}z$
2. Compute the differential: Determine that $dt=\frac{dz}{1+z^2}$
3. Rewrite the integral: Express the integral as $\int e^{t}dt$ using the substitution
4. Integrate: Evaluate the simpler integral to get $e^t+C$
5. Back-substitute: Replace $t$ with ${\tan }^{-1}z$ to obtain the final answer $e^{{\tan }^{-1}z}+C$