InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Evaluate:`intsqrt(2x^2+3x+4)dx` |
|
Answer» We have `(2x^(2)+3x+4)=2(x^(2)+(3)/(2)x+2)` `=2*{(x^(2)+(3)/(2)x+(9)/(16))+(2-(9)/(16))}=2*{(x+(3)/(4))^(2)+(sqrt(23)/(4))^(2)}`. `thereforesqrt(2x^(2)+3x+4)=sqrt(2)*sqrt((x+(3)/(4))^(2)+(sqrt(23)/(4))^(2))dx` `rArrintsqrt(2x^(2)+3x+4)dx=sqrt(2)*intsqrt((x+(3)/(4))^(2)+(sqrt(23)/(4))^(2))dx` `=sqrt(2)intsqrt(t^(2)+((sqrt(23))/(4))^(2))dt " where"(x+(3)/(4))=t` `=sqrt(2)*{(t)/(2)*sqrt(t^(2)+(23)/(16))+(23)/(32)log|t+sqrt(t^(2)+(23)/(16))|}+C` `[becauseintsqrt(t^(2)+a^(2))dt=(t)/(2)sqrt(t^(2)+a^(2))+(a^(2))/(2)log|t+sqrt(t^(2)+a^(2))|+C]` `=(sqrt(2))/(2)(x+(3)/(4))sqrt((x+(3)/(4))^(2)+(23)/(16))+(23sqrt(2))/(32)log|(x+(3)/(4))+sqrt(x^(2)+(3)/(2)x+2)|+C` `=((4x+3))/(4sqrt(2))*sqrt(x^(2)+(3)/(2)x+2)+(23sqrt(2))/(32)log|((4x+3))/(4)+(sqrt(2x^(2)+3x+4))/(sqrt(2))|+C` `=((4x+3)sqrt(2x^(2)+3x+4))/(8)+(23sqrt(2))/(32)log|((4x+3))/(4)+(sqrt(2x^(2)+3x+4))/(sqrt(2))|+C` |
|