Chapter 7 Class 12 Integrals
Chapter 7 Class 12 Integrals
Last updated at July 14, 2026 by Teachoo
Transcript
Ex 7.7, 11 ā«1āćā(š„2 ā8š„+7) " " ć šš„ 1/2 (x - 4)ā(š„2 ā8š„+7) + 9 log |š„ā4+ā(š„2ā8š„+7)| + C 1/2 (x + 4)ā(š„2 ā8š„+7) + 9 log |š„+4+ā(š„2ā8š„+7)| + C 1/2 (x - 4)ā(š„2 ā8š„+7) - 3ā2 log |š„ā4+ā(š„2ā8š„+7)| + C 1/2 (x - 4)ā(š„2 ā8š„+7) + 9/2 log |š„ā4+ā(š„2ā8š„+7)| + C ā«1āćā(š„^2ā8š„+7) šš„ć =ā«1āćā(š„^2ā2(4)(š„)+7) šš„ć =ā«1āćā(š„^2ā2(4)(š„) ć+(4)ć^2ā(4)^2+7) šš„ć =ā«1āćā((š„ā4)^2ā16+7) šš„ć =ā«1āćā((š„ā4)^2ā9 ) šš„ć =ā«1āćā((š„ā4)^2ā(3)^2 ) šš„ć =(š„ ā 4)/2 ā((š„ā4)^2ā(3)^2 )ā(3)^2/2 ššš|š„ā4+ā((š„ā4)^2ā(3)^2 )|+š¶ =(š„ ā 4)/2 ā(š„^2ā8š„+16ā9 )ā9/2 ššš|š„ā4+ā(š„^2ā8š„+16ā9)|+š¶ =(š„ ā 4)/2 ā(š„^2ā8š„+7)ā9/2 ššš|š„ā4+ā(š„^2ā8š„+7)|+š¶ ā“ Option (D) is correct. It is of the form ā«1āćā(š„^2āš^2 ) šš„=š„/2 ā(š„^2āš^2 )āš^2/2 ššš|š„+ā(š„^2āš^2 )|+š¶ć ā“ Replacing š„ by š„ā4 and a by 3 , we get