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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following:
$\int\frac{\text{x}}{\sqrt{\text{x}}+1}\text{dx}$
Hint: Put $\sqrt{\text{x}}=\text{z}$
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Answer

Let $\text{I}=\int\frac{\text{x}}{\sqrt{\text{x}}+1}\text{dx}$
Put $\sqrt{\text{x}}=\text{t}$
$\Rightarrow\ \frac{1}{2\sqrt{\text{x}}}\text{dx}=\text{dt}$
$\Rightarrow\ \text{dx}=2\sqrt{\text{x}}\text{dt}$
$\Rightarrow\ \text{dx}=2\text{t}\text{dt}$
Substituting $\sqrt{\text{x}}=\text{t}$ and dx = 2t in I, we get
$\text{I}=2\int\frac{\text{t}^2\cdot\text{t}}{\text{t}+1}\text{dt}$ $\Big[\because\sqrt{\text{x}}=\text{t}\Rightarrow\text{x}=\text{t}^2\Big]$
$=2\int\frac{\text{t}^3}{\text{t}+1}\text{dt}$
$=2\int\Big[(\text{t}^2-\text{t}+1)-\frac{1}{\text{t}+1}\Big]\text{dt}$
$=2\int(\text{t}^2-\text{t}+1)\text{dt}-2\int\frac{1}{\text{t}+1}\text{dt}$
$=2\Big[\frac{\text{t}^3}{3}-\frac{\text{t}^2}{3}+\text{t}-\log\big|(\text{t}-1)\big|\Big]+\text{C}$ $\bigg[\because\int\frac{1}{\text{x}}\text{dx}=\log|\text{x}|+\text{C and}\int\text{x}^\text{n}\text{dx}=\frac{\text{x}^{\text{n}+1}}{\text{n}}+\text{C}\bigg]$
$=2\bigg[\frac{\text{x}\sqrt{\text{x}}}{3}-\frac{\text{x}}{2}+\sqrt{\text{x}}-\log\big|(\sqrt{\text{x}}-1)\big|\bigg]+\text{C}$ $\Big[\because\ \text{t}=\sqrt{\text{x}}\Big]$

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