Evaluate the integral: \(\int^1_0\frac{x}{x^2+1}dx\)
Solution and Explanation
Let I=\(\int_{0}^{1} \frac{x}{ x^2+1},dx\)
Let x2+1=t⇒2x dx=dt
When x=0,t=1 and when x=1,t=2
∴\(\int_{0}^{1} \frac{x}{ x^2+1},dx\)=\(\frac 12\)\(∫^2_1\)\(\frac{dt}{t}\)
=\(\frac 12\)\([log|t|]^2_1\)
=\(\frac 12\)[log2-log1]
=\(\frac 12\)log2
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