If∫dx(x2−2x+10)2=A(tan−1(x−13)+f(x)x2−2x+10)+C
where C is a constant of integration, then
A = 154 and f(x) = 9(x – 1)2
A = 181 and f (x) = 3(x – 1)
A = 154 and f(x) = 3(x – 1)
A = 127 and f (x) = 9(x – 1)
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∫dx(x2−2x+10)2=∫dx((x−1)21+9)2
Let x – 1 = 3tanθ
dx = 3sec2θdθ
∴127∫cos2θdθ=154∫(1+cos2θ)dθ=154(θ+sin2θ2)
=154(tan−1(x−13)+3(x−1)x2−2x+10)+C