1 1+ dx= 1. _e(x+ )+C 2. _e( + )+C 3. 2 ^2 x 2 +C 4. 1 2 [x+ ( + ) ]+C

4. 1 2 [x+ ( + ) ]+C Solution: I= 1 1+ dx = + dx Numerator can be written as, =A( + )+B d( + ) dx =(A-B) +(A+B) A-B=0 and A+B=1 A=1 2 =B I= [1 2 ( + )+1 2 ( - ) ]dx + I=1 2 (1+ - + )dx I=1 2 [1+ ( + ) ]+C

1 1+ dx= 1. _e(x+ )+C 2. _e( + )+C 3. 2 ^2 x 2 +C 4. 1 2 [x+ ( + …

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Question

$\int\frac{1}{1+\tan\text{x}}\text{ dx}=$

$\log_\text{e}(\text{x}+\sin\text{x})+\text{C}$
$\log_\text{e}(\sin\text{x}+\cos\text{x})+\text{C}$
$2\sec^2\frac{\text{x}}{2}+\text{C}$
$\frac{1}{2}\big[\text{x}+\log(\sin\text{x}+\cos\text{x})\big]+\text{C}$

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Answer

$\frac{1}{2}\big[\text{x}+\log(\sin\text{x}+\cos\text{x})\big]+\text{C}$

Solution:
$\text{I}=\int\frac{1}{1+\tan\text{x}}\text{ dx}$
$=\int\frac{\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
Numerator can be written as,
$\cos\text{x}=\text{A}(\sin\text{x}+\cos\text{x})+\text{B}\frac{\text{d}(\sin\text{x}+\cos\text{x})}{\text{dx}}$
$\cos\text{x}=(\text{A}-\text{B})\sin\text{x}+(\text{A}+\text{B})\cos\text{x}$
$\text{A}-\text{B}=0$ and $\text{A}+\text{B}=1$
$\text{A}=\frac{1}{2}=\text{B}$
$\text{I}=\int\frac{\big[\frac{1}{2}(\sin\text{x}+\cos\text{x})+\frac{1}{2}(\cos\text{x}-\sin\text{x})\big]\text{dx}}{\sin\text{x}+\cos\text{x}}$
$\text{I}=\frac{1}{2}\int\Big(1+\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\text{I}=\frac{1}{2}\big[1+\int(\sin\text{x}+\cos\text{x})\big]+\text{C}$

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