Evaluate the integrals using substitution.
∫201x+4−x2dx.
∫201x+4−x2dx=∫201x+4−x2−(12)2+(12)2dx=∫2014+14−[(x2−x+14)]dx=∫2014+14−(x−12)2dx=∫201(√172)2−(x−12)2dx
=12.√172[log∣∣ ∣∣√172+x−12√172−x+12∣∣ ∣∣]20(∵∫dxa2−x2=12alog∣∣a+xa−x∣∣)=1√17[log∣∣∣√17+2x−1√17−2x+1∣∣∣]20=1√17[log∣∣∣√17+3√17−3−log∣∣∣√17−1√17+1∣∣∣∣∣∣]=1√17log∣∣∣√17+3√17−3+√17−1√17+1∣∣∣(∵logm−logn=logmn)
=1√17log∣∣∣(√17+3)(√17+1)(√17−3)(√17−1)∣∣∣=1√17log∣∣∣20+4√1720−4√17∣∣∣=1√17log∣∣∣5+√175−√17×5+√175+√17∣∣∣
To remove square root term from denominator, we multiply by its conjugate i.e.,5+√17 in both numerator and denominator.
=1√17log∣∣∣(5+√17)2(5)2−(√17)2∣∣∣[∵(a−b)(a+b)=a2−b2]=1√17log∣∣∣25+17+10√1725−17∣∣∣1√17log∣∣∣42+10√178∣∣∣=1√17log∣∣∣21+5√174∣∣∣