Q.Evaluate: $\int \dfrac{dx}{x\sqrt{x^4-1}}$ (Hint: Put $x^2=\sec\theta$)

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:

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

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

The Precise Statement

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

The Step-by-Step Method

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

A Second Example (with a constant factor)

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

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 $\log|g(x)|$ …