Question

Evaluate the integral: \(\int^1_0\frac{x}{x^2+1}dx\)

Answer 1

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