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3.3 Q-3

Question Statement

Evaluate the integral:

∫x24+x2,dx\int \frac{x^2}{4 + x^2} , dx

Background and Explanation

To solve this integral, we can use the technique of splitting the fraction into more manageable parts. This involves expressing the numerator in a way that simplifies the integral. In this case, we will rewrite x2x^2 as (4+x2)βˆ’4(4 + x^2) - 4 to split the expression into two terms. Recognizing the arctangent form for integrals will also be helpful.


Solution

  1. Rewrite the integrand:
    Start by splitting the numerator x2x^2 as follows:
x2=(4+x2)βˆ’4 x^2 = (4 + x^2) - 4

This gives us:

x24+x2=(4+x2)βˆ’44+x2=1βˆ’44+x2 \frac{x^2}{4 + x^2} = \frac{(4 + x^2) - 4}{4 + x^2} = 1 - \frac{4}{4 + x^2}

So, the integral becomes:

∫x24+x2,dx=∫(1βˆ’44+x2),dx \int \frac{x^2}{4 + x^2} , dx = \int \left(1 - \frac{4}{4 + x^2}\right) , dx
  1. Split the integral:
    Now, we can break the integral into two separate parts:
∫(1βˆ’44+x2),dx=∫1,dxβˆ’4∫14+x2,dx \int \left(1 - \frac{4}{4 + x^2}\right) , dx = \int 1 , dx - 4 \int \frac{1}{4 + x^2} , dx
  1. Solve the integrals:
    • The first integral is straightforward:
∫1,dx=x \int 1 , dx = x
  • The second integral is a standard arctangent integral. We recognize that:
∫14+x2,dx=12tanβ‘βˆ’1(x2) \int \frac{1}{4 + x^2} , dx = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right)

Substituting this back into our expression:

xβˆ’4[12tanβ‘βˆ’1(x2)] x - 4 \left[\frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right)\right]

Simplifying:

xβˆ’2tanβ‘βˆ’1(x2)+C x - 2 \tan^{-1}\left(\frac{x}{2}\right) + C

Key Formulas or Methods Used

  • Splitting the Fraction: The technique of breaking the integrand into simpler parts, i.e., rewriting x2x^2 as (4+x2)βˆ’4(4 + x^2) - 4.
  • Standard Integral of Arctangent: The formula:
∫dxa2+x2=1atanβ‘βˆ’1(xa)+C \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C

Summary of Steps

  1. Rewrite the integrand:
    x24+x2=1βˆ’44+x2\frac{x^2}{4 + x^2} = 1 - \frac{4}{4 + x^2}

  2. Split the integral into two parts:
    ∫1,dxβˆ’4∫14+x2,dx\int 1 , dx - 4 \int \frac{1}{4 + x^2} , dx

  3. Solve the integrals:

    • ∫1,dx=x\int 1 , dx = x
    • ∫14+x2,dx=12tanβ‘βˆ’1(x2)\int \frac{1}{4 + x^2} , dx = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right)
  4. Combine the results:
    xβˆ’2tanβ‘βˆ’1(x2)+Cx - 2 \tan^{-1}\left(\frac{x}{2}\right) + C