Mathematics

Solve :
$$ \int e^{x^2+\text{ln x}} . dx $$


SOLUTION
We have,
$$I=\int e^{x^2+ln\ x} dx$$
$$I=\int e^{x^2}. e^{ln\ x} dx$$
$$I=\int e^{x^2}. x dx$$

Let $$t=x^2$$
$$dt=2xdx$$

Therefore,
$$I=\dfrac{1}{2}\int e^{t}dt$$
$$I=\dfrac{1}{2}e^t+C$$

On putting the value of $$t$$, we get
$$I=\dfrac{1}{2}e^{x^2}+C$$

Hence, this is the answer.
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Subjective Medium Published on 17th 09, 2020
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