AQA Core 4 - Monday 10th June 2013 (AM) - Official Thread

\int \frac{1}{2y} dy = \int \frac{1}{3(x^{2}-1)} dx[br]\[br]From previous question you have the partial fractions[br]\[br][br]\Rightarrow \frac{1}{2}\ {ln(y)} = \frac{1}{3}\ \int \frac{1}{x-1}\ - \frac{1}{2(x+1)}\ dx [br][br]\Rightarrow \frac{1}{2}\ {ln(y)} = \frac{1}{6}\ ln{2(x-1)\over (x+1)} +C[br]\[br]sub in (3,1) gives C = 0[br]\[br]\Rightarrow \frac{1}{2}\ {ln(y)} = \frac{1}{6}\ ln{2(x-1)\over (x+1)}[br]\[br]Multiply through by 6[br]\[br]\Rightarrow 3{ln(y)} = ln{2(x-1)\over (x+1)} \Rightarrow \ln{y^{3}} = ln{2(x-1)\over (x+1)} [br]\[br]Take exponentials of both sides which gives[br]\[br]y^{3}= {2(x-1)\over (x+1)}[br][br]

[br]2x - 1 = A(x + 1)^{2} + B(x-1)(x-1) + C(x-1)[br]\[br]first let x = -1 to eliminate A and B, find C[br]then let x = 1 to eliminate B and C, find A[br]\[br]Sub in value for A and C and let x = 0, find B[br]