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Solve int(1)/((x-2)(x^(2)+x+1))dx.

Solve∫1(x−2)(x2+x+1) dx\int \frac{1}{(x-2)(x^{2}+x+1)}\,dx.

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Evaluate definite integrals using the chain ruleright chevron icon

Full solution

Q. Solve∫1(x−2)(x2+x+1) dx\int \frac{1}{(x-2)(x^{2}+x+1)}\,dx.
  1. Decompose integrand into partial fractions: We need to decompose the integrand into partial fractions. To do this, we set up the equation:\newline1(x−2)(x2+x+1)=Ax−2+Bx+Cx2+x+1\frac{1}{(x-2)(x^2+x+1)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+x+1}\newlineWe will find the values of AA, BB, and CC that make this equation true.
  2. Clear fractions by multiplying: Multiply both sides of the equation by the denominator (x−2)(x2+x+1)(x-2)(x^2+x+1) to clear the fractions:\newline1=A(x2+x+1)+(Bx+C)(x−2)1 = A(x^2+x+1) + (Bx+C)(x-2)\newlineNow we will find AA, BB, and CC by comparing coefficients or plugging in suitable values for xx.
  3. Find values of A, B, and C: Let's plug in x=2x=2 to find AA, since that will eliminate the terms with BB and CC:1=A(22+2+1)1 = A(2^2+2+1)1=A(7)1 = A(7)A=17A = \frac{1}{7}
  4. Find AA by plugging in x=2x=2: To find BB and CC, we can either compare coefficients or choose other values for xx. Let's choose x=0x=0 to find CC:
    1=A(02+0+1)+C(0−2)1 = A(0^2+0+1) + C(0-2)
    1=(1/7)(1)−2C1 = (1/7)(1) - 2C
    1=1/7−2C1 = 1/7 - 2C
    x=2x=200
    x=2x=211
    x=2x=222
  5. Find C by plugging in x=0x=0: Now we need to find B. We can compare coefficients for the xx term. The coefficient of xx on the left side is 00, and on the right side, it is A+B−2CA + B - 2C. We already know AA and CC, so we can solve for BB:
    0=A+B−2C0 = A + B - 2C
    0=17+B+670 = \frac{1}{7} + B + \frac{6}{7}
    xx00
  6. Find BB by comparing coefficients: Now that we have AA, BB, and CC, we can write the integral as:\newline∫1(x−2)(x2+x+1)dx=∫171(x−2)dx+∫−x−371(x2+x+1)dx\int\frac{1}{(x-2)(x^2+x+1)}dx = \int\frac{1}{7}\frac{1}{(x-2)}dx + \int\frac{-x-3}{7}\frac{1}{(x^2+x+1)}dx\newlineWe will integrate each term separately.
  7. Write integral with A, B, and C: Integrate the first term: ∫17÷(x−2) dx=17ln⁡∣x−2∣+C1\int \frac{1}{7} \div (x-2)\,dx = \frac{1}{7}\ln|x-2| + C_1
  8. Integrate first term: For the second term, we need to complete the square for the denominator x2+x+1x^2+x+1:x2+x+1=(x+12)2+34x^2+x+1 = (x+\frac{1}{2})^2 + \frac{3}{4}Now we can rewrite the integral as:∫−x−37x2+x+1 dx=∫−x−37(x+12)2+34 dx\int\frac{-x-\frac{3}{7}}{x^2+x+1}\,dx = \int\frac{-x-\frac{3}{7}}{(x+\frac{1}{2})^2 + \frac{3}{4}}\,dx
  9. Complete the square for the denominator: We can split the second integral into two parts: \int\frac{-x-\frac{\(3\)}{\(7\)}}{(x+\frac{\(1\)}{\(2\)})^\(2\) + \frac{\(3\)}{\(4\)}}dx = \int\frac{-x}{(x+\frac{\(1\)}{\(2\)})^\(2\) + \frac{\(3\)}{\(4\)}}dx + \int\frac{-\frac{\(3\)}{\(7\)}}{(x+\frac{\(1\)}{\(2\)})^\(2\) + \frac{\(3\)}{\(4\)}}dx
  10. Split second integral into two parts: Integrate the first part of the second term using substitution:\(\newlineLet u=x+12u = x + \frac{1}{2}, then du=dxdu = dx, and the integral becomes:\newline∫−uu2+34du\int \frac{-u}{u^2 + \frac{3}{4}}du\newlineThis is the integral of the form −uu2+a2\frac{-u}{u^2 + a^2}, which can be integrated as −12ln⁡(u2+a2)+C2-\frac{1}{2} \ln(u^2 + a^2) + C_2.
  11. Integrate first part using substitution: Integrate the second part of the second term using a trigonometric substitution:\newlineLet u=x+12u = x + \frac{1}{2}, and a2=34a^2 = \frac{3}{4}. Then the integral becomes:\newline∫−37(1u2+(3/2)2)du\int \frac{-3}{7}\left(\frac{1}{u^2 + (\sqrt{3}/2)^2}\right)du\newlineThis is the integral of the form au2+a2\frac{a}{u^2 + a^2}, which can be integrated as −aarctan⁡(ua)+C3-a \arctan(\frac{u}{a}) + C_3.
  12. Integrate second part using trigonometric substitution: Combine the results from steps 77, 1010, and 1111: \int \frac{\(1\)}{(x\(-2\))(x^\(2\)+x+\(1\))}\,dx = \frac{\(1\)}{\(7\)}\ln|x\(-2| - \frac{11}{22} \ln\left((x+\frac{11}{22})^22 + \frac{33}{44}\right) - \frac{33}{77} \arctan\left(\frac{22}{\sqrt{33}}(x+\frac{11}{22})\right) + C
  13. Combine results from steps 77, 1010, and 1111: Simplify the final answer by combining the constants:\newline∫1(x−2)(x2+x+1)dx=17ln⁡∣x−2∣−12ln⁡((x+12)2+34)−37arctan⁡(23(x+12))+C\int \frac{1}{(x-2)(x^2+x+1)}dx = \frac{1}{7}\ln|x-2| - \frac{1}{2} \ln\left((x+\frac{1}{2})^2 + \frac{3}{4}\right) - \frac{3}{7} \arctan\left(\frac{2}{\sqrt{3}}(x+\frac{1}{2})\right) + C\newlineWhere CC is the constant of integration.

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