How do you integrate $\int x {(4 x + 5)}^{3}$ $intx(4x+5)^3$ using substitution?

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How do you integrate $\int x {(4 x + 5)}^{3}$ $intx(4x+5)^3$ using substitution?

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Answer 1 · 2017-02-07T18:05:16

$\Rightarrow \frac{1}{80} {(4 x + 5)}^{5} - \frac{5}{64} {(4 x + 5)}^{4} + C$ $=>1/80(4x + 5)^5 - 5/64(4x + 5)^4 + C$

Explanation:

Let $u = 4 x + 5$ $u = 4x + 5$ . Then $d u = 4 d x \to d x = \frac{d u}{4}$ $du =4dx -> dx= (du)/4$ . This also means that $x = \frac{u - 5}{4}$ $x = (u - 5)/4$ .

$\Rightarrow \int \frac{u - 5}{4} u^{3} \frac{d u}{4}$ $=>int (u - 5)/4 u^3 (du)/4$

$\Rightarrow \frac{1}{16} \int u^{4} - 5 u^{3} d u$ $=>1/16intu^4 - 5u^3du$

$\Rightarrow \frac{1}{16} (\frac{1}{5} u^{5} - \frac{5}{4} u^{4}) + C$ $=>1/16(1/5u^5 - 5/4u^4) + C$

$\Rightarrow \frac{1}{80} u^{5} - \frac{5}{64} u^{4} + C$ $=>1/80u^5 - 5/64u^4 + C$

$\Rightarrow \frac{1}{80} {(4 x + 5)}^{5} - \frac{5}{64} {(4 x + 5)}^{4} + C$ $=>1/80(4x + 5)^5 - 5/64(4x + 5)^4 + C$

Hopefully this helps!

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How do you integrate $\int x {(4 x + 5)}^{3}$ $intx(4x+5)^3$ using substitution?

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How do you integrate $\int x {(4 x + 5)}^{3}$ $intx(4x+5)^3$ using substitution?