Find the anti-derivative (using substitution)?

Question

Find the anti-derivative (using substitution)?

$\int x^{3} {(x^{2} + 1)}^{99}$ $int x^3(x^2+1)^99$
Find the antiderivative of this using the substitution method.
I do know (and understand) that u = $x^{2} + 1$ $x^2+1$ and therefore
$d u =$ $du =$ $2 x$ $2x$ ( $\frac{d u}{2} = x$ $(du)/2=x$ ). However, I don't understand that even if there's an $x^{3}$ $x^3$ , you just plug it in normally, instead of making $u^{3}$ $u^3$ when you put it in.

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Answer 1 · 2018-03-10T07:57:18

$\int x^{3} {(x^{2} + 1)}^{99} d x = \frac{1}{20200} (10 x - 1) (10 x + 1) {(x^{2} + 1)}^{100} + C$ $intx^3(x^2+1)^99dx=1/20200(10x-1)(10x+1)(x^2+1)^100+C$

Explanation:

Let

$I = \int x^{3} {(x^{2} + 1)}^{99} d x$ $I=intx^3(x^2+1)^99dx$

Apply the substitution $u = x^{2} + 1$ $u=x^2+1$ :

$I = \int {(u - 1)}^{\frac{3}{2}} \cdot u^{99} \cdot \frac{d u}{2 \sqrt{u - 1}}$ $I=int(u-1)^(3/2)* u^99*(du)/(2sqrt(u-1))$

Simplify:

$I = \frac{1}{2} \int (u^{100} - u^{99}) d u$ $I=1/2int(u^100-u^99)du$

Integrate directly:

$I = \frac{1}{2} (\frac{1}{101} u^{101} - \frac{1}{100} u^{100}) + C$ $I=1/2(1/101u^101-1/100u^100)+C$

Reverse the substitution and simplify:

$I = \frac{1}{20200} (10 x - 1) (10 x + 1) {(x^{2} + 1)}^{100} + C$ $I=1/20200(10x-1)(10x+1)(x^2+1)^100+C$

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Find the anti-derivative (using substitution)?

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