Integrate the following function: $\frac{e^{\tan^{-1} x}}{1+x^2}$

Integrate the following function: frace^tan⁻¹ x1+x²

Exercise 7.2 · Q18

\frac{e^{\tan^{-1} x}}{1+x^2}

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Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given $\cos(x^2) \cdot 2x$ , find the original function. That's what u substitution does — it reverses the chain rule.

The Core Intuition

When an integral looks like "a function times the derivative of its inside," substitute the inside with $u$ and the derivative of the inside with $du$ . Consider:

\int 2x \cos(x^2) \, dx

Here $2x$ is the derivative of $x^2$ , and $x^2$ is the inside of $\cos(x^2)$ . Let $u = x^2$ , so $du = 2x \, dx$ :

\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ .

The Precise Statement

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided $g$ is differentiable and the resulting integral in $u$ is simpler.

The Step-by-Step Method

Identify a function $g(x)$ whose derivative $g'(x)$ also appears (possibly up to a constant factor).
Set $u = g(x)$ , compute $du = g'(x) \, dx$ .
Rewrite the entire integral in $u$ and $du$ — every $x$ and $dx$ must be replaced.
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Watch out

You cannot mix variables. If any $x$ remains after substitution, you chose the wrong $u$ (or must solve for $x$ in terms of $u$ — rare).

A Second Example (with a constant factor)

Evaluate $\int x \sqrt{x^2 + 1} \, dx$ . Let $u = x^2 + 1$ , so $x \, dx = \frac{1}{2} du$ :

\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

$x \cdot f(x^2)$ — derivative of $x^2$ is $2x$ , so $u = x^2$
$e^{g(x)} \cdot g'(x)$ — derivative of $g(x)$ appears
$\frac{g'(x)}{g(x)}$ — leads to $\log|g(x)|$ …

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The integral simplifies via the substitution $u = \tan^{-1} x$ , because the denominator $1+x^2$ is exactly the derivative of $\tan^{-1} x$ . The result is $e^{\tan^{-1} x} + C$ .

Why This Approach Works

When you see an integral like $\frac{e^{\tan^{-1} x}}{1+x^2}$ , the key is to notice the structure: the numerator contains $e^{\tan^{-1} x}$ , and the denominator is $1+x^2$ . The derivative of $\tan^{-1} x$ is $\frac{1}{1+x^2}$ . This is a classic setup for U Substitution — you have a function and its derivative (up to a constant) appearing together. The exponential function $e^u$ is one of the easiest functions to integrate, so letting $u = \tan^{-1} x$ turns the messy expression into a clean exponential integral.

Tip

Whenever you see $e^{\text{something}}$ multiplied by the derivative of that "something", substitution is almost always the way. Here, the "something" is $\tan^{-1} x$ , and its derivative $\frac{1}{1+x^2}$ is right there.

Step-by-Step Solution

Set up the substitution. Let $u = \tan^{-1} x$ . Then differentiate:

\frac{du}{dx} = \frac{1}{1+x^2}

This implies $du = \frac{dx}{1+x^2}$ .

Rewrite the integral in terms of $u$ . The original integral is:

\int \frac{e^{\tan^{-1} x}}{1+x^2} \, dx

Substituting $u$ and $du$ gives:

$\int e^u \, du$ …

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Q.Integrate the following function: $\frac{e^{\tan^{-1} x}}{1+x^2}$

Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The Core Intuition

The Precise Statement

The Step-by-Step Method

A Second Example (with a constant factor)

When Does It Work?

Why This Approach Works

Step-by-Step Solution

Method: Substitute an inverse-trig exponent

Steps

Common Mistakes

Q.Integrate the following function: etan⁡−1x1+x2\frac{e^{\tan^{-1} x}}{1+x^2}1+x2etan−1x​

Why This Approach Works

Step-by-Step Solution

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Q.Integrate the following function: $\frac{e^{\tan^{-1} x}}{1+x^2}$

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