Skip to content
Question of 373

Q.Prove that ∫ √(x² + a²) dx = (x/2) √(x² + a²) + (a²/2) log |x + √(x² + a²)| + c.

Goa GbshseGBSHSE Class 12 Board Exam 2018Subjective· 3mImportance★★★★★
0% · 0/373 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 →

Integrate by parts (treating x2+a2\sqrt{x^2+a^2} as 1⋅x2+a21\cdot\sqrt{x^2+a^2}), then solve for II algebraically.

Let I=∫x2+a2 dxI = \displaystyle\int\sqrt{x^2+a^2}\,dx. Integrating by parts with u=x2+a2u=\sqrt{x^2+a^2}, dv=dxdv=dx:

I=xx2+a2−∫x⋅xx2+a2 dx=xx2+a2−∫x2x2+a2 dxI = x\sqrt{x^2+a^2} - \displaystyle\int x\cdot\dfrac{x}{\sqrt{x^2+a^2}}\,dx = x\sqrt{x^2+a^2} - \displaystyle\int\dfrac{x^2}{\sqrt{x^2+a^2}}\,dx

Write x2=(x2+a2)−a2x^2 = (x^2+a^2)-a^2:

I=xx2+a2−∫x2+a2 dx+a2∫dxx2+a2I = x\sqrt{x^2+a^2} - \displaystyle\int\sqrt{x^2+a^2}\,dx + a^2\displaystyle\int\dfrac{dx}{\sqrt{x^2+a^2}}

I=xx2+a2−I+a2log⁡∣x+x2+a2∣I = x\sqrt{x^2+a^2} - I + a^2\log\left|x+\sqrt{x^2+a^2}\right|

…

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.