How do you find the integral $\int {(5 x + 4)}^{4} d x$ $int(5x+4)^4dx$ using substitution?

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Answer 1 · 2017-10-05T15:52:44

$= \frac{1}{25} {(5 x + 4)}^{5} + C$ $=1/25(5x+4)^5+C$

By substitution

$\int {(5 x + 4)}^{4} d x$ $int(5x+4)^4dx$

$u = 5 x + 1 \Rightarrow d u = 5 d x \Rightarrow d x = \frac{d u}{5}$ $u=5x+1=>du=5dx=>dx=(du)/5$

$\int {(5 x + 4)}^{4} d x = \int u^{4} \cdot \frac{d u}{5} = d x$ $int(5x+4)^4dx=intu^4*(du)/5=dx$

$= \frac{1}{5} \int u^{4} d u = \frac{1}{5} \times \frac{1}{5} u^{5} + C$ $=1/5intu^4du=1/5xx1/5u^5+C$

$= \frac{1}{25} u^{5} + C = \frac{1}{25} {(5 x + 4)}^{5} + C$ $=1/25u^5+C=1/25(5x+4)^5+C$

How do you find the integral ∫(5x+4)4dxint(5x+4)^4dx using substitution?