Question
Evalute the following integrals:
$\int\frac{1+\tan\text{x}}{\text{x}+\log\sec\text{x}}\text{dx}$
$\int\frac{1+\tan\text{x}}{\text{x}+\log\sec\text{x}}\text{dx}$
✓
Answer
Note: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{1+\tan\text{x}}{\text{x}+\log\sec\text{x}}\text{dx}$
Putting $\text{x}+\log\sec\text{x}=\text{t}$
$\Rightarrow1+\frac{\sec\text{x}\tan\text{x}}{\sec\text{c}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\tan\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{x}+\log\sec\text{x}|+\text{C}\ \big[\because\text{t}=\text{x}+\log\sec\text{x}\big]$
Let $\text{I}=\int\frac{1+\tan\text{x}}{\text{x}+\log\sec\text{x}}\text{dx}$
Putting $\text{x}+\log\sec\text{x}=\text{t}$
$\Rightarrow1+\frac{\sec\text{x}\tan\text{x}}{\sec\text{c}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\tan\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{x}+\log\sec\text{x}|+\text{C}\ \big[\because\text{t}=\text{x}+\log\sec\text{x}\big]$
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