Question
Find $\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$
✓
Answer
We have $\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$ = $\int \frac{\left(1-\frac{1}{x^{3}}\right)^{\frac{1}{4}}}{x^{4}} d x$
Put, $1-\frac{1}{x^{3}}=1-x^{-3}=t$ $\Rightarrow \frac{3}{x^{4}} d x=d t$
Therefore $\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$ = $\frac{1}{3} \int t^{\frac{1}{4}} d t=\frac{1}{3} \times \frac{4}{5} t^{\frac{5}{4}}+C=\frac{4}{15}\left(1-\frac{1}{x^{3}}\right)^{\frac{5}{4}}+C$
Put, $1-\frac{1}{x^{3}}=1-x^{-3}=t$ $\Rightarrow \frac{3}{x^{4}} d x=d t$
Therefore $\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$ = $\frac{1}{3} \int t^{\frac{1}{4}} d t=\frac{1}{3} \times \frac{4}{5} t^{\frac{5}{4}}+C=\frac{4}{15}\left(1-\frac{1}{x^{3}}\right)^{\frac{5}{4}}+C$
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