How do you integrate $\int \frac{3 x^{2}}{\sqrt{x^{3} + 7}}$ $int (3x^2)/sqrt(x^3+7)$ using substitution?

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Answer 1 · 2016-09-12T19:06:47

$2 \sqrt{x^{3} + 7} + C$ $2sqrt(x^3+7)+C$

$\int \frac{3 x^{2}}{\sqrt{x^{3} + 7}} d x$ $int(3x^2)/sqrt(x^3+7)dx$

Apply the substitution $x^{3} + 7 = u$ $x^3+7=u$ . Thus, $3 x^{2} d x = d u$ $3x^2dx=du$ . Substituting in:

$= \int \frac{d u}{\sqrt{u}}$ $=int(du)/sqrtu$

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

Using the rule $\int u^{n} d u = \frac{u^{n + 1}}{n + 1} + C$ $intu^ndu=u^(n+1)/(n+1)+C$ , where $n \neq - 1$ $n!=-1$ :

$= \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C$ $=u^(1/2)/(1/2)+C$

$= 2 \sqrt{u} + C$ $=2sqrtu+C$

$= 2 \sqrt{x^{3} + 7} + C$ $=2sqrt(x^3+7)+C$

How do you integrate ∫3x2√x3+7int (3x^2)/sqrt(x^3+7) using substitution?