Evaluate: $\int \dfrac{x^2}{1-x^4}\,dx$ (put $x^2=t$)

Evaluate: ∫ (x²)/(1-x⁴)dx (put x²=t)

NCERT Exemplar · Q19

\int \dfrac{x^2}{1-x^4}\,dx

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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Split $\dfrac{x^2}{1-x^4}=\dfrac12\left(\dfrac{1}{1-x^2}-\dfrac{1}{1+x^2}\right)$ and integrate each part: the result is $\dfrac14\log\left|\dfrac{1+x}{1-x}\right|-\dfrac12\tan^{-1}x+C$ .

Setting up. The suggested move $x^2=t$ replaces $x^2$ by $t$ but leaves an $x$ in $dx=\tfrac{dt}{2x}$ , so it does not fully rationalise the integral. The clean route uses the same factorisation it points to — $1-x^4=(1-x^2)(1+x^2)$ — to split the fraction directly.

1. Split the integrand

Write the numerator as a difference of the two factors of the denominator:

x^2=\frac12\big[(1+x^2)-(1-x^2)\big].

Then

\frac{x^2}{1-x^4}=\frac{\tfrac12[(1+x^2)-(1-x^2)]}{(1-x^2)(1+x^2)}=\frac12\left(\frac{1}{1-x^2}-\frac{1}{1+x^2}\right).

(Quick check: $\dfrac{1}{1-x^2}-\dfrac{1}{1+x^2}=\dfrac{2x^2}{1-x^4}$ , and half of that is $\dfrac{x^2}{1-x^4}$ . ✓)

2. Integrate term by term

\int\frac{x^2}{1-x^4}\,dx=\frac12\int\frac{dx}{1-x^2}-\frac12\int\frac{dx}{1+x^2}.

Use the standard results …

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Q.Evaluate: $\int \dfrac{x^2}{1-x^4}\,dx$ (put $x^2=t$ )

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?

1. Split the integrand

2. Integrate term by term

Method: Substituting $t=x^2$ then splitting into partial fractions

Steps

Common Mistakes

Q.Evaluate: ∫x21−x4 dx\int \dfrac{x^2}{1-x^4}\,dx∫1−x4x2​dx (put x2=tx^2=tx2=t)

1. Split the integrand

2. Integrate term by term

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Q.Evaluate: $\int \dfrac{x^2}{1-x^4}\,dx$ (put $x^2=t$ )

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