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Evaluate the integrals in Exercises 1 to 8 using substitution.

    Q7.    \int_{-1}^1\frac{dx}{x^2 +2x + 5}

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\int_{-1}^1\frac{dx}{x^2 +2x + 5}
the Dr can be written as x^2+2x+5 = (x+1)^2+2^2
and put x+1 = t then dx =dt

when x= -1 then t = 0 and when x = 1 then t = 2

\\\Rightarrow \int_{0}^{2}\frac{dt}{t^2+2^2}\\ =\frac{1}{2}[\tan^{-1}\frac{t}{2}]^2_0\\ =\frac{1}{2}( \pi/4)\\ =\frac{\pi}{8}

Posted by

manish

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