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♪ Integration by sustitution
∫ (e^tan-¹x)/1+x² dx
Put tan-¹x = t
Differentiate w.r.t t on both sides..
=> 1/1+x² dx = dt
=> ∫ e^t dt
=> e^t + C { :• ∫ e^x dx = e^x + C
Now again substitute the value of t
=> e^tan-¹x +C
∫ (e^tan-¹x)/1+x² dx
Put tan-¹x = t
Differentiate w.r.t t on both sides..
=> 1/1+x² dx = dt
=> ∫ e^t dt
=> e^t + C { :• ∫ e^x dx = e^x + C
Now again substitute the value of t
=> e^tan-¹x +C
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