dfrac√(x)1+√(x)

Exercise 11.6 · Q10

\dfrac{\sqrt{x}}{1+\sqrt{x}}

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Substituting $u=\sqrt x$ turns the surd into a polynomial-over-linear fraction that simplifies by ordinary division.

Step 1. Substitute. Let $u=\sqrt x$ , so $x=u^2$ , $dx=2u\,du$ .

Step 2. Rewrite. $\displaystyle\int\dfrac{\sqrt x}{1+\sqrt x}dx=\int\dfrac{u}{1+u}\cdot2u\,du=2\int\dfrac{u^2}{1+u}du$ .

Step 3. Divide the improper fraction. $\dfrac{u^2}{1+u}=u-1+\dfrac1{1+u}$ (since $u^2=(u+1)(u-1)+1$ ).

Step 4. Integrate. $2\displaystyle\int\left(u-1+\dfrac1{1+u}\right)du=2\left(\dfrac{u^2}2-u+\log|1+u|\right)+c=u^2-2u+2\log|1+u|+c$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CA Foundation 2025Set jan-20251 markMCQ
$\displaystyle\int (2x + 5)^7 \, dx$
›Reveal solutionSolution

$\int (2x+5)^7\,dx = \dfrac{(2x+5)^8}{2 \cdot 8} = \dfrac{(2x+5)^8}{16} + C$ .

Step 1 — Use the standard form for a linear substitution
$\int (ax+b)^n\, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$
Step 2 — Identify the constants
$a = 2, \quad b = 5, \quad n = 7$
Step 3 — Substitute
$\int (2x+5)^7\, dx = \frac{(2x+5)^{8}}{2 \times 8} = \frac{(2x+5)^8}{16} + C$
Why the other options are wrong: (B) divides only by 7 (raises to the wrong power AND drops the $\frac{1}{a}$ ); (C) and (D) wrongly expand $(2x+5)$ as $(2x^2+5x)$ , which is not the inner function. …
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CA Foundation 2024Set sep-20241 markMCQ
$\int \dfrac{1}{x(x^5 + 1)} \, dx$
›Reveal solutionSolution

Substitute u = x⁵ then use partial fractions: the integral = (1/5)·ln(x⁵/(x⁵+1)) + c.

Step 1 — Prepare for substitution

Multiply top and bottom by $x^4$ :
$\int \dfrac{1}{x(x^5+1)}\,dx = \int \dfrac{x^4}{x^5(x^5+1)}\,dx$
Step 2 — Substitute u = x⁵

With $u = x^5$ , $du = 5x^4\,dx$ , i.e. $x^4\,dx = \dfrac{du}{5}$ :
$= \dfrac{1}{5}\int \dfrac{du}{u(u+1)}$
Step 3 — Partial fractions and integrate
$\dfrac{1}{u(u+1)} = \dfrac{1}{u} - \dfrac{1}{u+1}$ $\dfrac{1}{5}\int\!\left(\dfrac{1}{u} - \dfrac{1}{u+1}\right)du = \dfrac{1}{5}\big[\ln u - \ln(u+1)\big] + c = \dfrac{1}{5}\ln\!\left(\dfrac{x^5}{x^5+1}\right) + c$
Why the other options are wrong: (A) omits the 1/5 factor; (C) uses 1/3 (wrong power); (D) inverts the ratio to (x⁵+1)/x⁵. Only (B) has both the 1/5 and the correct x⁵/(x⁵+1). …
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Q. $\dfrac{\sqrt{x}}{1+\sqrt{x}}$

Concept understanding — Integration by Substitution

Step 1 — Use the standard form for a linear substitution

Step 2 — Identify the constants

Step 3 — Substitute

Step 1 — Prepare for substitution

Step 2 — Substitute u = x⁵

Step 3 — Partial fractions and integrate

Q.x1+x\dfrac{\sqrt{x}}{1+\sqrt{x}}1+x​x​​

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Q. $\dfrac{\sqrt{x}}{1+\sqrt{x}}$

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