How do you use substitution to integrate $x^{2} \sqrt{x + 1}$ $x^2sqrt(x+1)$ ?

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How do you use substitution to integrate $x^{2} \sqrt{x + 1}$ $x^2sqrt(x+1)$ ?

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Answer 1 · 2015-09-25T05:18:36

Refer to explanation

Explanation:

We set $u = \sqrt{x + 1}$ $u=sqrt(x+1)$ hence $d u = \frac{1}{2 \cdot \sqrt{x + 1}} d x$ $du=1/(2*sqrt(x+1))dx$

$\int x^{2} \cdot \sqrt{x + 1} d x = 2 \cdot \int {(u^{2} - 1)}^{2} \cdot u^{2} d u = 2 \cdot \int (u^{6} - 2 u^{4} + u^{2}) d u = 2 \cdot [\frac{u^{7}}{7} - 2 \cdot \frac{u^{5}}{5} + \frac{u^{3}}{3}]$ $int x^2*sqrt(x+1)dx=2*int (u^2-1)^2*u^2du= 2*int (u^6-2u^4+u^2)du=2*[u^7/7-2*u^5/5+u^3/3]$

Now replacing u with $\sqrt{x + 1}$ $sqrt(x+1)$ we get finally that

$\int x^{2} \cdot \sqrt{x + 1} d x = \frac{2}{105} \cdot \sqrt{{(x + 1)}^{3}} \cdot (15 x^{2} - 12 x + 8) + c$ $int x^2*sqrt(x+1)dx=2/105*sqrt((x+1)^3)*(15x^2-12x+8)+c$

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How do you use substitution to integrate $x^{2} \sqrt{x + 1}$ $x^2sqrt(x+1)$ ?

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How do you use substitution to integrate $x^{2} \sqrt{x + 1}$ $x^2sqrt(x+1)$ ?