A-Level Techniques of integration: worked solution

Question

Given 3x²/[(x+1)(3x²+x+1)] = 1/(x+1) − 1/(3x²+x+1), integrate 3x²/[(x+1)(3x²+x+1)] with respect to x. [4]

Use the identity supplied in the question to split the integral into two easier ones: ∫3x²/[(x+1)(3x²+x+1)] dx = ∫1/(x+1) dx − ∫1/(3x²+x+1) dx The first integral is the standard logarithm form: ∫1/(x+1) dx = ln|x+1| For the second, the denominator does not factorise (its discriminant is 1² − 4(3)(1) = −11 < 0), so complete the square. Take out the factor 3 first: 3x² + x + 1 = 3(x² + x/3) + 1 = 3[(x + 1/6)² − 1/36] + 1 = 3(x + 1/6)² − 1/12 + 1 = 3(x + 1/6)² + 11/12 So ∫1/(3x²+x+1) dx = ∫ dx/[3(x + 1/6)² + 11/12] = (1/3)∫ dx/[(x + 1/6)² + 11/36] This is now the standard form ∫du/(u² + k²) = (1/k)arctan(u/k), with u = x + 1/6 and k² = 11/36, so k = √11/6: = (1/3) × (6/√11) arctan((x + 1/6)/(√11/6)) = (2/√11) arctan((6x + 1)/√11) (the inner fraction was multiplied top and bottom by 6). Putting the two parts together: ∫3x²/[(x+1)(3x²+x+1)] dx = ln|x+1| − (2/√11) arctan((6x+1)/√11) + C