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Evaluate the integrals in Exercises 1 to 8 using substitution.

    Q1.    \int_0^1\frac{x}{x^2 +1}dx

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\int_0^1\frac{x}{x^2 +1}dx
let x^2+1 = t \Rightarrow xdx =dt/2
when x = 0 then t = 1 and when x =1 then t = 2
\therefore \int_{o}^{1}\frac{x}{x^2+1}dx=\frac{1}{2}\int_{1}^{2}\frac{dt}{t}
                                    \\=\frac{1}{2}[\log\left | t \right |]_{1}^{2}\\ =\frac{1}{2}\log 2

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manish

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