How do you integrate $\int 36 x^{3} {(3 x^{4} + 3)}^{5}$ $int36x^3(3x^4+3)^5$ using substitution?

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Answer 1 · 2016-12-07T10:38:32

The answer is $= \frac{729}{2} {(x^{4} + 1)}^{6} + C$ $=729/2(x^4+1)^6+C$

We use $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} \cdot C (n \neq - 1)$ $intx^ndx=x^(n+1)/(n+1)*C (n!=-1)$

We do this by substitution

let $u = 3 (x^{4} + 1)$ $u=3(x^4+1)$

$d u = 3 \cdot 4 x^{3} d x = 12 x^{3} d x$ $du=3*4x^3dx=12x^3dx$

$x^{3} d x = \frac{d u}{12}$ $x^3dx=(du)/12$

$\int 36 x^{3} {(3 x^{4} + 5)}^{5} d x$ $int36x^3(3x^4+5)^5dx$

$= \frac{36}{12} \int u^{5} d u$ $=36/12intu^5du$

$= 3 \frac{u^{6}}{6}$ $=3u^6/6$

$= \frac{1}{2} u^{6} = \frac{1}{2} (3^{6} {(x^{4} + 1)}^{6}) + C$ $=1/2u^6=1/2(3^6(x^4+1)^6)+C$

$= \frac{729}{2} {(x^{4} + 1)}^{6} + C$ $=729/2(x^4+1)^6+C$

How do you integrate ∫36x3(3x4+3)5int36x^3(3x^4+3)^5 using substitution?