Skip to content
3.1 · Q3

Q.Find:

(i) ∫1−2x1+x2 dx\int \frac{1-2x}{\sqrt{1+x^2}}\,dx
(ii) ∫13x2+2x−1 dx\int \frac{1}{\sqrt{3x^2+2x-1}}\,dx
Yanam CbseNCERTSubjective· 3mImportance★★★★★
66% · 39/59 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Split (i) into a standard form plus a substitution; complete the square in (ii) to reach a ∫dxx2−a2\int\frac{dx}{\sqrt{x^2-a^2}} form.

∫dxx2+1=log⁡∣x+x2+1∣+C\displaystyle\int\frac{dx}{\sqrt{x^2+1}}=\log\big|x+\sqrt{x^2+1}\big|+C; ∫dxx2−a2=log⁡∣x+x2−a2∣+C\displaystyle\int\frac{dx}{\sqrt{x^2-a^2}}=\log\big|x+\sqrt{x^2-a^2}\big|+C.

(i) ∫1−2x1+x2 dx\displaystyle\int \frac{1-2x}{\sqrt{1+x^2}}\,dx

  1. Split: ∫dx1+x2−∫2x1+x2 dx.\displaystyle\int\frac{dx}{\sqrt{1+x^2}}-\int\frac{2x}{\sqrt{1+x^2}}\,dx.
  2. First term =log⁡∣x+1+x2∣.=\log\big|x+\sqrt{1+x^2}\big|.
  3. Second: put u=1+x2⇒du=2x dx⇒∫u−1/2du=2u=21+x2.u=1+x^2\Rightarrow du=2x\,dx\Rightarrow \int u^{-1/2}du=2\sqrt{u}=2\sqrt{1+x^2}.
  4. =log⁡∣x+1+x2∣−21+x2+C.=\log\big|x+\sqrt{1+x^2}\big|-2\sqrt{1+x^2}+C.

(ii) ∫13x2+2x−1 dx\displaystyle\int \frac{1}{\sqrt{3x^2+2x-1}}\,dx

5. Factor 33: 3x2+2x−1=3(x2+23x−13).3x^2+2x-1=3\big(x^2+\tfrac23x-\tfrac13\big).

6. Complete square: x2+23x=(x+13)2−19x^2+\tfrac23x=\big(x+\tfrac13\big)^2-\tfrac19, so x2+23x−13=(x+13)2−19−13=(x+13)2−49.x^2+\tfrac23x-\tfrac13=\big(x+\tfrac13\big)^2-\tfrac19-\tfrac13=\big(x+\tfrac13\big)^2-\tfrac49. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.