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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following $\int \large \frac{( x)}{\sqrt x+1}$$dx \quad(Hint:Put\;\sqrt x=z)$

$\begin{array}{1 1} (A) \; 2 \bigg[\large\frac{x \sqrt x}{3}-\frac{x}{2}-\sqrt x-log |\sqrt x+1|\bigg]+c \\ (B) \;2 \bigg[\large\frac{x \sqrt x}{3}+\frac{x}{2}+\sqrt x-log |\sqrt x+1|\bigg]+c \\ (C) \; 2 \bigg[\large\frac{x \sqrt x}{3}-\frac{x}{2}+\sqrt x+log |\sqrt x+1|\bigg]+c \\ (D)\;2 \bigg[\large\frac{x \sqrt x}{3}-\frac{x}{2}+\sqrt x-log |\sqrt x+1|\bigg]+c \end{array} $

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1 Answer

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Toolbox:
  • $a^3+b^3=(a+b)(a^2-ab+b^2)$
  • $\int \large\frac{dx}{(x+a)}$=$ log |x+a|+c$
  • $\int x^n dx=\large\frac{x^{n+1}}{n+1}+c$
Step 1:
Let $I=\large\int\frac{ x}{\sqrt x+1} dx$
Let $\sqrt x=z => x=z^2$
differentiating with respect to z
$\large \frac{1}{2 \sqrt x}dx =dz$
$dx=2 \sqrt x dz$
$=2zdz$
Substituting this in I we get
$I=\Large \int \frac{z^2.(2zdz)}{z+1}$
$=2 \Large\int \frac{z^3dz}{z+1}$
Add and subtract 1 in the numerator
$I=2 \int \Large\frac{z^3+1-1}{z+1}$ $ dz$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
Similarly, $(z^3+1)=(z+1)(z^2-z+1)$
Hence $I=2 \int \bigg[\large\frac{(z+1)(z^2-z+1)}{z+1} \bigg]-\bigg(\frac{1}{z+1}\bigg) dz$
$= 2 \bigg \{ \int (z^2-z+1)dz -\int \large\frac{1}{z+1}\bigg\} dz$
Step 2:
On integrating we get
$I=2 \bigg \{ \bigg[ \Large\frac{z^3}{3}-\frac{z^2}{2}+z\bigg]$-$log |z+1| \bigg\}+c$
substituting for z we get
$ I= 2 \bigg\{\bigg[\Large\frac{\sqrt x)^3}{3}-\frac{(\sqrt x)^2}{2}+\sqrt x\bigg]$-$log |\sqrt x+1|\bigg\}+c$
$ I= 2 \bigg[\Large\frac{x \sqrt x}{3}-\frac{x}{2}$+$\sqrt x-log |\sqrt x+1|\bigg]+c$
answered Apr 11, 2013 by meena.p
edited Apr 11, 2013 by meena.p
 
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