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$$ \int \sqrt{x}\left(1+\sqrt{x}\right)^3dx $$ I tried to solve this by using substitution. However, I could not reach the answer. I tried to replace $1+\sqrt{x} = u$ and $\sqrt{x} = u$ But still did not get the answer.I don't want to open parenthesis and solve integral in that way.

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  • $\begingroup$ Do you mean $$\int \sqrt{x} \cdot \left(1+\sqrt{x}\right)^3\;?$$ $\endgroup$ – Chinnapparaj R Dec 17 '18 at 15:35
  • $\begingroup$ Yes,thanks for correcting $\endgroup$ – Arif Rustamov Dec 17 '18 at 15:40
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    $\begingroup$ Mutliply out the cube to get four terms. Multiply through by the square root. Then you have an easy sum of powers. $\endgroup$ – Ethan Bolker Dec 17 '18 at 15:40
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    $\begingroup$ Why exactly would you not want to expand the integrand? $\endgroup$ – KM101 Dec 17 '18 at 15:42
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    $\begingroup$ If you do a substitution you will still have parentheses that need opening: you may get three rather than four terms, but at the cost of the substitution, and for an indefinite integral the cost of re-substitution $\endgroup$ – Henry Dec 17 '18 at 15:55
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Let $u = \sqrt{x}+1 \implies \mathrm{d}x = 2\sqrt{x}\mathrm{d}u$. It will then be $x = (u-1)^2$. Thus, the integral becomes :

$$\int (u-1)^2u^3\mathrm{d}u ={\displaystyle\int}u^5\,\mathrm{d}u-\class{steps-node}{\cssId{steps-node-2}{2}}{\displaystyle\int}u^4\,\mathrm{d}u+{\displaystyle\int}u^3\,\mathrm{d}u =\dfrac{u^6}{6}-\dfrac{2u^5}{5}+\dfrac{u^4}{4} + C $$

Now, substitute for $u= \sqrt{x} + 1$ and you should get for the initial integral :

\begin{align*} \int \sqrt{x}(1+\sqrt{x}) \mathrm{d}x &=\dfrac{\left(\sqrt{x}+1\right)^6}{3}-\dfrac{4\left(\sqrt{x}+1\right)^5}{5}+\dfrac{\left(\sqrt{x}+1\right)^4}{2} + C\\ &= \boxed{\dfrac{\left(\sqrt{x}+1\right)^4\left(10x-4\sqrt{x}+1\right)}{30}+C}. \end{align*}

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  • $\begingroup$ Can we try $ x^{\frac{-1}{2}}+1=u $ by taking $\sqrt{x}$ common $\endgroup$ – Lakshya Sinha Dec 17 '18 at 16:05
  • $\begingroup$ @LakshyaSinha What? $\endgroup$ – Rebellos Dec 17 '18 at 16:05
  • $\begingroup$ Take $ \sqrt{x} $ common from cubic term and suppose that as u $\endgroup$ – Lakshya Sinha Dec 17 '18 at 16:07
  • $\begingroup$ The person who disliked would be kind enough to point out ? The solution is perfectly clear and correct. $\endgroup$ – Rebellos Dec 18 '18 at 18:42
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By parts $u=(1+\sqrt{x})^3$ and $v'=\sqrt{x}$:

$$\int \sqrt{x}\left(1+\sqrt{x}\right)^3dx=\frac{2}{3}x^{\frac{3}{2}}\left(1+\sqrt{x}\right)^3-\int \:x\left(1+\sqrt{x}\right)^2dx$$

And then by expanding: $$\int \:x\left(1+\sqrt{x}\right)^2dx=\frac{x^2}{2}+\frac{4}{5}x^{\frac{5}{2}}+\frac{x^3}{3}$$

So $$\int \sqrt{x}\left(1+\sqrt{x}\right)^3dx=\frac{2}{3}x^{\frac{3}{2}}\left(1+\sqrt{x}\right)^3-\frac{x^2}{2}-\frac{4}{5}x^{\frac{5}{2}}-\frac{x^3}{3}+C.$$

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