How do you integrate by substitution $\int \frac{x^{3}}{\sqrt{1 + x^{4}}} d x$ $int x^3/sqrt(1+x^4) dx$ ?

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How do you integrate by substitution $\int \frac{x^{3}}{\sqrt{1 + x^{4}}} d x$ $int x^3/sqrt(1+x^4) dx$ ?

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Answer 1 · 2016-12-19T01:52:18

$\int \frac{x^{3}}{\sqrt{1 + x^{4}}} d x = \frac{\sqrt{1 + x^{4}}}{2} + C$ $intx^3/sqrt(1+x^4)dx=sqrt(1+x^4)/2+C$

Explanation:

Using integration by substitution

let $u = 1 + x^{4} \Rightarrow d u = 4 x^{3} d x$ $u = 1+x^4 => du = 4x^3dx$

Then, together with the known formula $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx = x^(n+1)/(n+1)+C$ for $n \neq - 1$ $n!=-1$ , we have

$\int \frac{x^{3}}{\sqrt{1 + x^{4}}} d x = \frac{1}{4} \int \frac{4 x^{3}}{\sqrt{1 + x^{4}}} d x$ $intx^3/sqrt(1+x^4)dx = 1/4int(4x^3)/sqrt(1+x^4)dx$

$= \frac{1}{4} \int \frac{1}{\sqrt{u}} d u$ $=1/4int1/sqrt(u)du$

$= \frac{1}{4} \int u^{- \frac{1}{2}} d u$ $=1/4intu^(-1/2)du$

$= \frac{1}{4} (\frac{u^{\frac{1}{2}}}{\frac{1}{2}}) + C$ $=1/4(u^(1/2)/(1/2))+C$

$= \frac{u^{\frac{1}{2}}}{2} + C$ $=u^(1/2)/2+C$

$= \frac{\sqrt{1 + x^{4}}}{2} + C$ $=sqrt(1+x^4)/2+C$

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How do you integrate by substitution $\int \frac{x^{3}}{\sqrt{1 + x^{4}}} d x$ $int x^3/sqrt(1+x^4) dx$ ?

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Explanation:

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How do you integrate by substitution $\int \frac{x^{3}}{\sqrt{1 + x^{4}}} d x$ $int x^3/sqrt(1+x^4) dx$ ?