Write down the constants \(A\) and \(B\) such that, for all values of \(x\),\n\(\n1-x=A(2x-4)+B.\n\)\nHence find \(\displaystyle \int \frac{1-x}{x^2-4x+1}\,dx\).[4]
Hence, evaluate \(\displaystyle \int_{\frac{1}{2}}^{1}\frac{1}{x^3}e^{\frac{1}{x}}\,dx\), giving your answer in terms of \(e\).[3]
By using the substitution \(x=3\sec\theta\) for \(0\leq \theta \leq \pi\), find\n\(\n\int_{3}^{6}\frac{\sqrt{x^2-9}}{x}\,dx,\n\)\nleaving your answer in the form \(a\sqrt{b}+c\pi\), where \(a,b,c\) are constants to be found.[5]
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Answer:(a) \(A=-\dfrac12,\ B=-1\), integral as shown (b)(i) \(-x^{-2}e^{1/x}\) (b)(ii) \(e^2\) (c) \(3\sqrt3-\pi\)