Mathematics

$$\int \dfrac{\sqrt [2] {x}}{1+\sqrt [4] {x^{3}}}dx$$ is equal to


SOLUTION
We have,
$$I=\int \dfrac{\sqrt [2] {x}}{1+\sqrt [4] {x^{3}}}dx$$

Let 
$$t=1+x^{\dfrac{3}{4}}$$
$$\dfrac{dt}{dx}=0+\dfrac{3}{4}x^{\dfrac{-1}{4}}$$

$$dt=\dfrac{3}{4x^{\dfrac{1}{4}}}dx$$

$$dx=\dfrac{4x^{\dfrac{1}{4}}}{3}dt$$

Therefore,
$$I=\dfrac{4}{3}\displaystyle \int \dfrac{t-1}{t}dt$$
$$I=\dfrac{4}{3}\displaystyle \int \left( 1-\dfrac{1}{t}\right)dt$$
$$I=\dfrac{4}{3}\displaystyle \left( t-\ln t\right)+C$$

On putting the value of $$t$$, we get
$$I=\dfrac{4}{3}\displaystyle \left( 1+x^{\frac{3}{4}}-\ln (1+x^{\frac{3}{4}})\right)+C$$

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