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Question : 72 of 160
Marks:
+1,
-0
Solution:
I=∫dx It is solved as
I=∫| (2x2+6√2x+9)−(3√2x+) |
| 2x2+6√2x+9 |
dx =∫dx−∫dx =−∫dx Now
3√2x+=A(4x+6√2)+B Compare the coefficients of like terms.
4A=3√2 A= And
6√2A+B= B=− Therefore
3√2x+=3(4x+6√2)+ Hence.
I=−∫dx =−3∫dx+∫dx =−log|2x2+6√2x+9|+∫dx =−log|(√2x+3)2|+⋅+c′ Solve further
I=−log|(√2x+3)|++c′ =[(√2x+3)−6log|√2x+3|−]+c
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