Mathematics

$$\int {{e^{3{{\log }_e}x}}.{{\left( {{x^4} + 1} \right)}^{ - 1}}dx = \_\_\_\_\_\_\_\_\_ + C.} $$


ANSWER

$$\frac{1}{4}\log \left( {{x^4} + 1} \right)$$


SOLUTION
$$ I=\int { { e^{ 3\log_ex } }.{ { \left( { { x^{ 4 } }+1 } \right)  }^{ -1 } }dx }  $$
$$ I=\int { { e^{ \log_ex^3 } }.{ { \left( { { x^{ 4 } }+1 } \right)  }^{ -1 } }dx }  $$
$$ I=\int \dfrac{ { x^3  }}{{ { \left( { { x^{ 4 } }+1 } \right)  } }}dx  $$

Let $$t=x^4+1$$
$$dt=4x^3dx$$

Therefore,
$$I=\dfrac{1}{4}\int \dfrac{1}{t}dt$$
$$I=\dfrac{1}{4}\log(t)+C$$

Put the value of $$t$$, we get
$$I=\dfrac{1}{4}\log(x^4+1)+C$$

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