How do you integrate $\int (2 x^{\frac{3}{2}} + 1) \sqrt{x}$ $int (2x^(3/2)+1)sqrtx$ using substitution?

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How do you integrate $\int (2 x^{\frac{3}{2}} + 1) \sqrt{x}$ $int (2x^(3/2)+1)sqrtx$ using substitution?

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Answer 1 · 2018-03-29T03:18:05

$\int (2 x^{\frac{3}{2}} + 1) \sqrt{x} d x = \frac{1}{4} {(2 x^{\frac{3}{2}} + 1)}^{2} + C$ $int(2x^(3/2)+1)sqrtxdx=1/4(2x^(3/2)+1)^2+C$

Explanation:

Our goal with substitution is to find an expression whose differential appears in the integral, and represent it using the variable $u .$ $u.$

We have $\int (2 x^{\frac{3}{2}} + 1) \sqrt{x} d x$ $int(2x^(3/2)+1)sqrtxdx$

In this case, $u = 2 x^{\frac{3}{2}} + 1$ $u=2x^(3/2)+1$ is the best possible choice. Were we to choose $u = \sqrt{x}, d u = \frac{1}{2 \sqrt{x}} d x$ $u=sqrtx, du=1/(2sqrtx)dx$ , and this differential is not even close to anything in the integral.

Calculate its differential:

$d u = (2) (\frac{3}{2}) x^{\frac{1}{2}} d x$ $du=(2)(3/2)x^(1/2)dx$

$d u = 2 \sqrt{x} d x$ $du=2sqrtxdx$

Divide both sides by $2 :$ $2:$

$\frac{1}{2} d u = \sqrt{x} d x$ $1/2du=sqrtxdx$

And we see $\sqrt{x} d x$ $sqrtxdx$ does in fact show up in the integral. So, rewrite with the substitution:

$\frac{1}{2} \int u d u = \frac{1}{2} (\frac{1}{2}) u^{2} + C = \frac{1}{4} u^{2} + C$ $1/2intudu=1/2(1/2)u^2+C=1/4u^2+C$

Rewriting in terms of $x$ $x$ yields

$\int (2 x^{\frac{3}{2}} + 1) \sqrt{x} d x = \frac{1}{4} {(2 x^{\frac{3}{2}} + 1)}^{2} + C$ $int(2x^(3/2)+1)sqrtxdx=1/4(2x^(3/2)+1)^2+C$

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How do you integrate $\int (2 x^{\frac{3}{2}} + 1) \sqrt{x}$ $int (2x^(3/2)+1)sqrtx$ using substitution?

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Explanation:

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