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Integration : $\displaystyle\int 2x(x^2+2)^{\frac32}\ dx$

...... $$=\int 2x(x^2+2)^{\frac32}\ dx$$ $$=2\int x(x^2+2)^{^{\frac32}}\ dx$$ Let $x^2=y$. Now, $$=2\int \sqrt {y} (y+2)^{^{\frac32}}\ dy$$ ....

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    $\begingroup$ How about $$\sqrt{x^2+2}=u\implies x^2+2=u^2$$ $\endgroup$ – lab bhattacharjee May 12 '17 at 14:52
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    $\begingroup$ Just let $u = x^{2} + 2$ from the start. Then your integration amounts to finding the primitive of $u^{3/2}$. $\endgroup$ – mattos May 12 '17 at 14:53
  • $\begingroup$ You have forgotten the $dx$. Without this, a substitution will give the wrong answer. $\endgroup$ – Fly by Night May 12 '17 at 15:11
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Let $x^2+2=z^2,\implies2xdx=2zdz$ and $\left(x^2+2\right)^{3/2}=z^3$.

So, \begin{align*} \int 2x(x^2+2)^{\frac32}\ dx&=\int z^3\cdot2z\ dz\\ &=2\int z^4\ dz\\ &=\dfrac{2}{5}z^5+c\hspace{25pt}\text{[where, c=integration const.]}\\ &=\dfrac{2}{5}\left(x^2+2\right)^{\frac52}+c \end{align*} Or, without assuming anything you can do as:

$$\int2x(x^2+2)^{\frac32}\ dx=\int(x^2+2)^{\frac32}\ d(x^2+2)=\dfrac{1}{1+\frac32}(x^2+2)^{1+\frac32}+c=\dfrac{2}{5}\left(x^2+2\right)^{\frac52}+c$$

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What everyone is implicitly saying is that you've made a mistake. You're first equality is fine, but in your second one, because you've not been writing "$dx$" or "$dy$", you've forgotten to divide by the derivative. So you let $x^2 = y$, but then didn't then say "$dy = 2 x dx$". As such, you're supposed to have $$ \int 2x (x^2 + 3)^{3/2} dx = \int 2x (x^2 + 3)^{3/2} \frac1{2x} dy = \int (y+3)^{3/2} dy. $$ You can then use the normal rules to say that this final integral equals $$ \tfrac25(y+3)^{5/2} + \text{constant} = \tfrac25 (x^2 + 3)^{5/2} + \text{constant}. $$

While the other answers are 100% correct, hopefully this enlightens you as to your mistake :)

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$$y = x^2 +2$$ $$dy = 2x \ dx$$

$$I = \int dy \ y^\frac{3}{2} = \frac{2}{5} y^\frac{5}{2} +C = (x^2 +2)^\frac{5}{2} +C $$

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  • $\begingroup$ How is $\textrm {dy}=2x \textrm {dx}$? $\endgroup$ – Aryabhatta May 12 '17 at 15:01
  • $\begingroup$ @Aryabhatta $dy/dx = 2x$ $\endgroup$ – Spine Feast May 12 '17 at 15:02
  • $\begingroup$ @SpineFast, How is $$I=\int dy y^{\dfrac {3}{2}}=\dfrac {2}{5} y^{\dfrac {5}{2}} + C$$? $\endgroup$ – Aryabhatta May 12 '17 at 15:05
  • $\begingroup$ @Aryabhatta $\int dy \ y^r = y^{r+1}/(r+1) +C$ $\endgroup$ – Spine Feast May 12 '17 at 15:31
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Hint:

use the substitution $$ x^2+2=u \qquad \Rightarrow \qquad 2xdx=du $$

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