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Question : 74 of 160
Marks:
+1,
-0
Solution:
I=∫dx =∫dx =∫dx+∫dx Let,
I=I1+I2 I1=∫dx =∫| x4−x2+1 |
| (1+x2)(x4−x2+1) |
dx =∫dx=tan−1x I2=∫dx Let
x3=t⇒x2dx= I2=‌∫=tan−1(t) =tan−1(x3 ) Therefore,
I=tan−1x+tan−1(x3)+C
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