Using the substitution \(x=\tan\theta\), find the exact value of \(\displaystyle \int_0^1 \frac{1}{\left(x^2+1\right)^2}\, dx.\)[5]
Find the exact value of \(\displaystyle \int_0^1 \frac{P(x)}{\left(x^2+1\right)^2\left(x^2+4\right)}\, dx\) if
\(P(x)=\left(x^2+1\right)\left(x^2+4\right),\)[1]
\(P(x)=\left(x^2+1\right)^2.\)[1]
By considering\n\(\nx^4+3x^2+14=a\left(x^2+1\right)^2+b\left(x^2+1\right)\left(x^2+4\right)+c\left(x^2+4\right),\n\)\nfind the exact value of\n\(\n\int_0^1 \frac{x^4+3x^2+14}{\left(x^2+1\right)^2\left(x^2+4\right)}\, dx,\n\)\nleaving your answer in the form \(p+q\pi+\tan^{-1} r\), where \(p\), \(q\) and \(r\) are rational numbers to be determined.[5]
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Answer:\(\dfrac14+\dfrac\pi8\); final integral \(=1+\dfrac\pi4+\tan^{-1}(\dfrac12)\)