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Question: ∫(x^4)/((x^2)+1)...

∫(x^4)/((x^2)+1)

Answer

(x^3)/3 - x + arctan(x) + C

Explanation

Solution

To evaluate the integral ∫x4x2+1dx\int \frac{x^4}{x^2+1} dx, we first observe that the degree of the numerator (4) is greater than the degree of the denominator (2). In such cases, we perform polynomial long division.

Step 1: Perform Polynomial Long Division Divide x4x^4 by x2+1x^2+1: x4x2+1\frac{x^4}{x^2+1} We can rewrite x4x^4 as x2(x2+1)−x2x^2(x^2+1) - x^2: x2(x2+1)−x2x2+1=x2−x2x2+1\frac{x^2(x^2+1) - x^2}{x^2+1} = x^2 - \frac{x^2}{x^2+1} Now, we deal with the remaining fraction x2x2+1\frac{x^2}{x^2+1}. We can rewrite x2x^2 as (x2+1)−1(x^2+1) - 1: (x2+1)−1x2+1=1−1x2+1\frac{(x^2+1) - 1}{x^2+1} = 1 - \frac{1}{x^2+1} Substitute this back into the expression: x4x2+1=x2−(1−1x2+1)\frac{x^4}{x^2+1} = x^2 - \left(1 - \frac{1}{x^2+1}\right) x4x2+1=x2−1+1x2+1\frac{x^4}{x^2+1} = x^2 - 1 + \frac{1}{x^2+1}

Step 2: Integrate the Simplified Expression Now, integrate each term: ∫x4x2+1dx=∫(x2−1+1x2+1)dx\int \frac{x^4}{x^2+1} dx = \int \left(x^2 - 1 + \frac{1}{x^2+1}\right) dx We can split this into three separate integrals: ∫x2dx−∫1dx+∫1x2+1dx\int x^2 dx - \int 1 dx + \int \frac{1}{x^2+1} dx Using standard integration formulas:

  1. ∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C
  2. ∫1x2+a2dx=1atan⁡−1(xa)+C\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C

Applying these formulas:

  1. ∫x2dx=x2+12+1=x33\int x^2 dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}
  2. ∫1dx=x\int 1 dx = x
  3. ∫1x2+1dx=∫1x2+12dx=11tan⁡−1(x1)=tan⁡−1(x)\int \frac{1}{x^2+1} dx = \int \frac{1}{x^2+1^2} dx = \frac{1}{1} \tan^{-1}\left(\frac{x}{1}\right) = \tan^{-1}(x)

Combining these results and adding the constant of integration CC: ∫x4x2+1dx=x33−x+tan⁡−1(x)+C\int \frac{x^4}{x^2+1} dx = \frac{x^3}{3} - x + \tan^{-1}(x) + C

The final answer is x33−x+tan⁡−1(x)+C\frac{x^3}{3} - x + \tan^{-1}(x) + C.

Explanation:

  1. The integrand's numerator degree is higher than the denominator's, so polynomial long division is performed.
  2. The expression is simplified to x2−1+1x2+1x^2 - 1 + \frac{1}{x^2+1}.
  3. Each term is integrated using standard integration formulas: ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} and ∫1x2+a2dx=1atan⁡−1(xa)\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}).
  4. The results are combined, and the constant of integration CC is added.