How do you integrate $\int x {(4 x - 1)}^{4}$ $intx(4x-1)^4$ using substitution?

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How do you integrate $\int x {(4 x - 1)}^{4}$ $intx(4x-1)^4$ using substitution?

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Answer 1 · 2017-02-13T06:34:24

$\frac{1}{96} {(4 x - 1)}^{6} + \frac{1}{80} {(4 x - 1)}^{5} + C$ $1/96(4x-1)^6 + 1/80(4x-1)^5+C$

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The gist is to force the substitution to happen. We need to find the derivatives of the parts of the integrals so that we can substitute the right things. For example we "make" a $(32 x - 8)$ $(32x-8)$ term because it is the derivative of $(16 x^{2} - 8 x + 1)$ $(16x^2-8x+1)$ It is a lot like completing the square in quadratics.

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How do you integrate $\int x {(4 x - 1)}^{4}$ $intx(4x-1)^4$ using substitution?

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How do you integrate ∫x(4x−1)4intx(4x-1)^4 using substitution?

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How do you integrate $\int x {(4 x - 1)}^{4}$ $intx(4x-1)^4$ using substitution?