Evaluate: $\int_{0}^{1} \dfrac{x\,dx}{\sqrt{1+x^2}}$

Evaluate: ∫₀¹ dfracxdx√(1+x²)

NCERT Exemplar · Q32

\int_{0}^{1} \dfrac{x\,dx}{\sqrt{1+x^2}}

Punjab PsebShort· 3mImportance★★★★★

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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 $u = 1 + x^2$ , turning the integrand into $\frac{1}{2\sqrt{u}}$ . The value is $\boxed{\sqrt{2} - 1}$ .

Why U‑substitution works here

When you see a function and its derivative lurking in an integral, substitution is the natural move. In $\frac{x}{\sqrt{1+x^2}}$ , the numerator $x$ is (up to a constant) the derivative of $1+x^2$ . That’s the classic signal: let $u$ be the “inside” of the square root, and the $x\,dx$ will become a clean $\frac{1}{2}\,du$ .

The square root in the denominator suggests the antiderivative will involve something like $\sqrt{u}$ , and indeed it will.

Step‑by‑step

Choose the substitution.

Let $u = 1 + x^2$ . Then $du = 2x\,dx$ , so $x\,dx = \frac{1}{2}\,du$ .
Change the limits.

When $x = 0$ , $u = 1 + 0^2 = 1$ .

When $x = 1$ , $u = 1 + 1^2 = 2$ .
Rewrite the integral.

The original integral becomes

\int_{x=0}^{1} \frac{x\,dx}{\sqrt{1+x^2}} = \int_{u=1}^{2} \frac{\frac{1}{2}\,du}{\sqrt{u}} = \frac{1}{2} \int_{1}^{2} u^{-1/2}\,du.

Integrate.

\frac{1}{2} \int_{1}^{2} u^{-1/2}\,du = \frac{1}{2} \left[ \frac{u^{1/2}}{1/2} \right]_{1}^{2} = \frac{1}{2} \cdot 2 \left[ \sqrt{u} \right]_{1}^{2} = \left[ \sqrt{u} \right]_{1}^{2}.

Evaluate. $\sqrt{2} - \sqrt{1} = \sqrt{2} - 1.$ …

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Q.Evaluate: $\int_{0}^{1} \dfrac{x\,dx}{\sqrt{1+x^2}}$

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 U‑substitution works here

Step‑by‑step

Method: Substituting $u=1+x^{2}$ for $\int\frac{x\,dx}{\sqrt{1+x^2}}$

Steps

Common Mistakes

Q.Evaluate: ∫01x dx1+x2\int_{0}^{1} \dfrac{x\,dx}{\sqrt{1+x^2}}∫01​1+x2​xdx​

Why U‑substitution works here

Step‑by‑step

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Q.Evaluate: $\int_{0}^{1} \dfrac{x\,dx}{\sqrt{1+x^2}}$

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