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