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Question

Mathematics Question on integral

Evaluate the integral: ∫−11dxx2+2x+5∫_{-1}^1\frac {dx}{x^2+2x+5}

Answer

∫−11dxx2+2x+5∫_{-1}^1\frac {dx}{x^2+2x+5}= ∫−11dx(x2+2x+1)+4∫_{-1}^1\frac {dx}{(x^2+2x+1)+4} = ∫−11dx(x+1)2+22∫_{-1}^1\frac {dx}{(x+1)^2+2^2}

Let x+1=t  ⟹  dx=dtLet\ x+1=t \implies dx=dt

When x=−1,t=0 and whenx=1,t=2When\ x=-1,t=0\ and\ when x=1,t=2

∴∫−11dx(x+1)2+22∫_{-1}^1\frac {dx}{(x+1)^2+2^2} = ∫02dtt2+22∫_0^2 \frac {dt}{t^2+2^2}

                             =$[\frac 12 tan^{-1}\frac t2]_0^2$

                             =$\frac 12 tan^{-1}1-\frac 12 tan^{-1}0$

                             =$\frac 12(\frac \pi4)$

                             =$\frac\pi8$