How do you integrate $\int {(e^{4 x} - 4)}^{\frac{1}{5}} \cdot 8 e^{4 x}$ $int(e^(4x)-4)^(1/5)*8e^(4x)$ using substitution?

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How do you integrate $\int {(e^{4 x} - 4)}^{\frac{1}{5}} \cdot 8 e^{4 x}$ $int(e^(4x)-4)^(1/5)*8e^(4x)$ using substitution?

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Answer 1 · 2018-04-09T11:47:05

$I = \frac{5}{3} (e^{4 x} - 4) {(e^{4 x} - 4)}^{\frac{1}{5}} + c$ $I=5/3(e^(4x)-4)(e^(4x)-4)^(1/5)+c$

Explanation:

Here.

$I = \int {(e^{4 x} - 4)}^{\frac{1}{5}} \cdot 8 e^{4 x} d x = 2 \int {(e^{4 x} - 4)}^{\frac{1}{5}} 4 e^{4 x} d x$ $I=int(e^(4x)-4)^(1/5)*8e^(4x)dx=2int(e^(4x)-4)^(1/5)4e^(4x)dx$

Let,

$e^{4 x} - 4 = u^{5} \Rightarrow 4 e^{4 x} d x = 5 u^{4} d u$ $e^(4x)-4=u^5=>4e^(4x)dx=5u^4du$

$I = 2 \int {(u^{5})}^{\frac{1}{5}} 5 u^{4} d u$ $I=2int(u^5)^(1/5)5u^4du$

$= 10 \int u \cdot u^{4} d u$ $=10intu*u^4du$

$= 10 \int u^{5} d u$ $=10intu^5du$

$= 10 (\frac{u^{6}}{6}) + c$ $=10(u^6/6)+c$

$= \frac{5}{3} \cdot u^{5} \cdot u + c$ $=5/3*u^5*u+c$

$I = \frac{5}{3} (e^{4 x} - 4) {(e^{4 x} - 4)}^{\frac{1}{5}} + c$ $I=5/3(e^(4x)-4)(e^(4x)-4)^(1/5)+c$

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How do you integrate $\int {(e^{4 x} - 4)}^{\frac{1}{5}} \cdot 8 e^{4 x}$ $int(e^(4x)-4)^(1/5)*8e^(4x)$ using substitution?

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