How do you integrate $x^{3} {((x^{4}) + 3)}^{5} d x$ $x^3((x^4)+3)^5 dx$ ?

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How do you integrate $x^{3} {((x^{4}) + 3)}^{5} d x$ $x^3((x^4)+3)^5 dx$ ?

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Answer 1 · 2016-05-03T01:50:42

$\frac{{(x^{4} + 3)}^{6}}{24} + C$ $(x^4+3)^6/24+C$

Explanation:

We have the integral

$\int x^{3} {(x^{4} + 3)}^{5} d x$ $intx^3(x^4+3)^5dx$

This can be solved fairly simply through substitution: we don't have to go through the hassle of expanding the binomial and then integrating term by term.

Let $u = x^{4} + 3$ $u=x^4+3$ . This implies that $d u = 4 x^{3} d x$ $du=4x^3dx$ . Since we only have $x^{3} d x$ $x^3dx$ in the integral and not $4 x^{3} d x$ $4x^3dx$ , multiply the interior of the integral by $4$ $4$ and the exterior by $1 / 4$ $1//4$ to balance this.

$\int x^{3} {(x^{4} + 3)}^{5} d x = \frac{1}{4} \int {(x^{4} + 3)}^{5} \cdot 4 x^{3} d x = \frac{1}{4} \int u^{5} d u$ $intx^3(x^4+3)^5dx=1/4int(x^4+3)^5*4x^3dx=1/4intu^5du$

Integrating this results in $(\frac{1}{4}) \frac{u^{6}}{6} + C$ $(1/4)u^6/6+C$ , and since $u = x^{4} + 3$ $u=x^4+3$ , this becomes $\frac{{(x^{4} + 3)}^{6}}{24} + C$ $(x^4+3)^6/24+C$ .

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How do you integrate $x^{3} {((x^{4}) + 3)}^{5} d x$ $x^3((x^4)+3)^5 dx$ ?

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How do you integrate x3((x4)+3)5dx x^3((x^4)+3)^5 dx?

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How do you integrate $x^{3} {((x^{4}) + 3)}^{5} d x$ $x^3((x^4)+3)^5 dx$ ?