dfracx²1+x⁶

Exercise 11.6 · Q2

\dfrac{x^{2}}{1+x^{6}}

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✓ Free question

Recognising $x^6=(x^3)^2$ suggests the substitution $u=x^3$ , after which the integral becomes the standard $1/(1+u^2)$ form.

Step 1. Substitute. Let $u=x^3$ , so $du=3x^2dx$ , i.e. $x^2dx=\dfrac{du}3$ .

Step 2. Rewrite. $\displaystyle\int\dfrac{x^2dx}{1+x^6}=\dfrac13\int\dfrac{du}{1+u^2}$ .

Step 3. Integrate. $\dfrac13\tan^{-1}u+c$ .

Step 4. Re-substitute. $\dfrac13\tan^{-1}(x^3)+c$ .

Step 5. Check. $\dfrac{d}{dx}\left[\dfrac13\tan^{-1}(x^3)\right]=\dfrac13\cdot\dfrac{3x^2}{1+x^6}=\dfrac{x^2}{1+x^6}$ , matching the integrand.

✓Final answer

\dfrac{1}{3}\tan^{-1}(x^3)+c

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.

Watch out

Divide by BOTH the new power ( $n+1=8$ ) and the coefficient of $x$ ( $a=2$ ) → denominator $16$ . Missing the $\frac{1}{a}$ factor is the most common mistake.

Tip

Verify by differentiating: $\frac{d}{dx}\frac{(2x+5)^8}{16} = \frac{8(2x+5)^7 \cdot 2}{16} = (2x+5)^7$ . ✓

✓Final answer
$\dfrac{(2x+5)^8}{16}$
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).

Watch out

The constant is 1/5 (because du = 5x⁴dx introduces the 5), and the ratio inside the log is x⁵/(x⁵+1) — coming from ln u − ln(u+1), not the reverse.

Tip

For ∫1/[x(xⁿ+1)]dx, multiply by xⁿ⁻¹/xⁿ⁻¹ and set u = xⁿ; the answer is always (1/n)·ln(xⁿ/(xⁿ+1)) + c.

✓Final answer
$\dfrac{1}{5} \log \left(\dfrac{x^5}{x^5+1}\right) + c$

←Previous question · 11.7.4Q1 — dfracx√1+x² →Next question · 11.7.4Q3 — dfracex-e-xex+e-x

Q. $\dfrac{x^{2}}{1+x^{6}}$

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.x21+x6\dfrac{x^{2}}{1+x^{6}}1+x6x2​

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Q. $\dfrac{x^{2}}{1+x^{6}}$

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