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JEE Main 2021
Indefinite Integration
Indefinite Integrals
Hard

Question

If x+12x1dx\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx = f(x) 2x1\sqrt {2x - 1} + C, where C is a constant of integration, then f(x) is equal to :

Options

Solution

Key Concepts and Formulas

  • Indefinite Integration: The process of finding the antiderivative of a function.
  • Substitution Method: A technique to simplify integrals by substituting a part of the integrand with a new variable.
  • xndx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, where n1n \neq -1 and C is the constant of integration.

Step-by-Step Solution

Step 1: Perform a u-substitution. We are given the integral x+12x1dx\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx. To simplify this, we'll use the substitution method. Let t=2x1t = \sqrt{2x-1}. This substitution is chosen because it simplifies the square root in the denominator.

Step 2: Solve for x and find dx in terms of dt. Square both sides of the substitution equation to get t2=2x1t^2 = 2x - 1. Now, solve for x: 2x=t2+1x=t2+122x = t^2 + 1 \Rightarrow x = \frac{t^2 + 1}{2}. Next, differentiate both sides of t2=2x1t^2 = 2x - 1 with respect to x: 2tdtdx=22t \frac{dt}{dx} = 2. This implies dx=tdtdx = t \, dt.

Step 3: Substitute into the integral. Substitute x=t2+12x = \frac{t^2 + 1}{2} and dx=tdtdx = t \, dt into the original integral: x+12x1dx=t2+12+1ttdt=t2+1+22dt=t2+32dt\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx = \int {{{\frac{t^2 + 1}{2} + 1} \over t}} \, t \, dt = \int {{\frac{t^2 + 1 + 2}{2}}} \, dt = \int {{\frac{t^2 + 3}{2}}} \, dt.

Step 4: Evaluate the simplified integral. Now, integrate with respect to t: t2+32dt=12(t2+3)dt=12(t33+3t)+C=t36+3t2+C\int {{\frac{t^2 + 3}{2}}} \, dt = \frac{1}{2} \int (t^2 + 3) \, dt = \frac{1}{2} \left( \frac{t^3}{3} + 3t \right) + C = \frac{t^3}{6} + \frac{3t}{2} + C.

Step 5: Factor out t and simplify. Factor out t from the expression: t36+3t2+C=t6(t2+9)+C\frac{t^3}{6} + \frac{3t}{2} + C = \frac{t}{6} (t^2 + 9) + C.

Step 6: Substitute back for t. Substitute t=2x1t = \sqrt{2x-1} back into the expression: 2x16((2x1)2+9)+C=2x16(2x1+9)+C=2x16(2x+8)+C\frac{\sqrt{2x-1}}{6} ((\sqrt{2x-1})^2 + 9) + C = \frac{\sqrt{2x-1}}{6} (2x - 1 + 9) + C = \frac{\sqrt{2x-1}}{6} (2x + 8) + C.

Step 7: Simplify the expression. 2x16(2x+8)+C=2x13(x+4)+C\frac{\sqrt{2x-1}}{6} (2x + 8) + C = \frac{\sqrt{2x-1}}{3} (x + 4) + C.

Step 8: Identify f(x). We are given that the integral is equal to f(x)2x1+Cf(x) \sqrt{2x-1} + C. Comparing this with our result, we can identify f(x)=x+43f(x) = \frac{x+4}{3}.

Step 9: Rewrite f(x) to match the options. f(x)=13(x+4)=13x+43f(x) = \frac{1}{3}(x+4) = \frac{1}{3}x + \frac{4}{3}. The correct answer is 13(x+4)\frac{1}{3}(x+4). This corresponds to option (B).

Common Mistakes & Tips

  • Forgetting the constant of integration: Always remember to add the constant of integration, C, when evaluating indefinite integrals.
  • Incorrectly substituting back: Be careful when substituting back to the original variable. Ensure you have correctly solved for x in terms of t and substitute the right expression.
  • Algebraic errors: Double-check your algebraic manipulations to avoid mistakes.

Summary

We used the substitution method to evaluate the given indefinite integral. By substituting t=2x1t = \sqrt{2x-1}, we simplified the integral and were able to find its antiderivative. After substituting back to the original variable x, we identified the function f(x) as x+43\frac{x+4}{3}, which matches option (B).

Final Answer The final answer is \boxed{\frac{1}{3}(x+4)}, which corresponds to option (B).

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