Evaluate: $\int \dfrac{x}{1+\sqrt{x}}\,dx$ (Hint: Put $\sqrt{x}=z$)

> [!TLDR] > The integral $\int \frac{x}{1+\sqrt{x}}\,dx$ is solved by substituting $\sqrt{x}=z$, which rationalises the integrand into a simple rational function. After expanding and integrating term by term, the final answer is $\frac{2}{3}x^{3/2} - x + 2\sqrt{x} - 2\log(1+\sqrt{x}) + C$. The hint to put $\sqrt{x}=z$ is the key. Why? Because the integrand has a square root in the denominator, and substituting $z=\sqrt{x}$ turns every power of $x$ into a power of $z$, making the expression a plain rational function in $z$ — no radicals left. That’s the whole point: we trade a messy radical for a clean polynomial division. Let’s walk through it. 1. **Set up the substitution.** Let $\sqrt{x} = z$. Then $x = z^2$, and differentiating gives $dx = 2z\,dz$. The integral becomes: $$ \int \frac{x}{1+\sqrt{x}}\,dx = \int \frac{z^2}{1+z} \cdot 2z\,dz = 2\int \frac{z^3}{1+z}\,dz. $$ 2. **Simplify the rational function.** The integrand $\frac{z^3}{1+z}$ is an improper rational function (degree of numerator > degree of denominator). We must divide: $$ z^3 \div (z+1) = z^2 - z + 1 - \frac{1}{z+1}. $$ Check: $(z+1)(z^2 - z + 1) = z^3 + 1$, so indeed $$ \frac{z^3}{z+1} = z^2 - z + 1 - \frac{1}{z+1}. $$ This step is crucial — without it, we’d be stuck with a fraction that doesn’t integrate nicely. 3. **Integrate term by term.** Now the integral is: $$ 2\int \left( z^2 - z + 1 - \frac{1}{z+1} \right) dz. $$ Each term is elementary: $$ \begin{aligned} \int z^2\,dz &= \frac{z^3}{3}, \\ \int z\,dz &= \frac{z^2}{2}, \\ \int 1\,dz &= z, \\ \int \frac{1}{z+1}\,dz &= \log|z+1|. \end{aligned} $$ So: $$ 2\left( \frac{z^3}{3} - \frac{z^2}{2} + z - \log|z+1| \right) + C. $$ 4. **Back-substitute $z = \sqrt{x}$.** …

Evaluate: ∫ dfracx1+√(x)dx (Hint: Put √(x)=z)

NCERT Exemplar · Q10

\int \dfrac{x}{1+\sqrt{x}}\,dx

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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 $\int \frac{x}{1+\sqrt{x}}\,dx$ is solved by substituting $\sqrt{x}=z$ , which rationalises the integrand into a simple rational function. After expanding and integrating term by term, the final answer is $\frac{2}{3}x^{3/2} - x + 2\sqrt{x} - 2\log(1+\sqrt{x}) + C$ .

The hint to put $\sqrt{x}=z$ is the key. Why? Because the integrand has a square root in the denominator, and substituting $z=\sqrt{x}$ turns every power of $x$ into a power of $z$ , making the expression a plain rational function in $z$ — no radicals left. That’s the whole point: we trade a messy radical for a clean polynomial division.

Let’s walk through it.

Set up the substitution. Let $\sqrt{x} = z$ . Then $x = z^2$ , and differentiating gives $dx = 2z\,dz$ . The integral becomes:

\int \frac{x}{1+\sqrt{x}}\,dx = \int \frac{z^2}{1+z} \cdot 2z\,dz = 2\int \frac{z^3}{1+z}\,dz.

Simplify the rational function. The integrand $\frac{z^3}{1+z}$ is an improper rational function (degree of numerator > degree of denominator). We must divide:

z^3 \div (z+1) = z^2 - z + 1 - \frac{1}{z+1}.

Check: $(z+1)(z^2 - z + 1) = z^3 + 1$ , so indeed

\frac{z^3}{z+1} = z^2 - z + 1 - \frac{1}{z+1}.

This step is crucial — without it, we’d be stuck with a fraction that doesn’t integrate nicely.

Integrate term by term. Now the integral is:

2\int \left( z^2 - z + 1 - \frac{1}{z+1} \right) dz.

Each term is elementary:

\begin{aligned} \int z^2\,dz &= \frac{z^3}{3}, \\ \int z\,dz &= \frac{z^2}{2}, \\ \int 1\,dz &= z, \\ \int \frac{1}{z+1}\,dz &= \log|z+1|. \end{aligned}

So:

2\left( \frac{z^3}{3} - \frac{z^2}{2} + z - \log|z+1| \right) + C.

Back-substitute $z = \sqrt{x}$ . …

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Q.Evaluate: $\int \dfrac{x}{1+\sqrt{x}}\,dx$ (Hint: Put $\sqrt{x}=z$ )

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?

Method: Rationalising substitution for integrands with $\sqrt{x}$

Steps

Common Mistakes

Q.Evaluate: ∫x1+x dx\int \dfrac{x}{1+\sqrt{x}}\,dx∫1+x​x​dx (Hint: Put x=z\sqrt{x}=zx​=z)

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Q.Evaluate: $\int \dfrac{x}{1+\sqrt{x}}\,dx$ (Hint: Put $\sqrt{x}=z$ )

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