Consider using the method of partial fractions to solve x²+x+1 Sa 2x²+2+x²+x A In(x²+1) B x+2 D In(x²) x² +1 dx = [Aƒ (x) – B ln(g(x))] + D fb dx , and if A = }, B < 0, then g(x) = x+2

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A: Given integral is      (1) Its reduction is ∫Ax+Bx2+1+C(3x+2)+D9x2+12x+16dx We can write it as…

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