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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
$\int\frac{3\text{x}+5}{\sqrt{7\text{x}+9}}\text{dx}$
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Answer

$\text{Let I}=\int\Big(\frac{3\text{x}+5}{\sqrt{7\text{x}+9}}\Big)\text{dx}$
$\text{Putting}\ 7\text{x}+9=\text{t}$
$\Rightarrow\text{x}=\frac{\text{t}-9}{7}\ \&\ 7\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{7}$
$\therefore\text{I}=\int\Bigg(\frac{3\big(\frac{\text{t}-9}{7}\big)+5}{\sqrt{\text{t}}}\Bigg)\text{dt}$
$=\int\Big(\frac{3}{7}\frac{\text{t}}{\sqrt{\text{t}}}-\frac{27}{7\sqrt{\text{t}}}+\frac{5}{\sqrt{\text{t}}}\Big)\frac{\text{dt}}{7}$
$=\frac{3}{7\times7}\int\text{t}^\frac{1}{2}\text{dt}-\frac{27}{7\times7}\int\text{t}^{-\frac{1}{2}}\text{dt}+\frac{5}{7}\int^{-\frac{1}{2}}\text{dt}$
$=\frac{3}{7\times7}\bigg[\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg]-\frac{27}{7\times7}\bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg]+\frac{5}{7}\bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg]+\text{C}$
$=\frac{2}{7\times7}\text{t}^\frac{3}{2}-\frac{27}{7\times7}2\text{t}^\frac{1}{2}+\frac{10\sqrt{t}}{7}+\text{C}$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}-\frac{54}{7\times7}(7\text{x}+9)^\frac{1}{2}+\frac{10}{7}\sqrt{7\text{x}+9}+\text{C}$ $[\because\text{t}=7\text{x}+9]$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}+\Big(10-\frac{54}{7}\Big)\frac{\sqrt{7\text{x}+9}}{7}+\text{C}$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}+\Big(\frac{70-54}{7}\Big)\frac{\sqrt{7\text{x}+9}}{7}+\text{C}$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}+\frac{16}{7\times7}\frac{\sqrt{7\text{x}+9}}{7}+\text{C}$
$=\frac{2}{7\times7}\Big[(7\text{x}+9)^\frac{1}{2}[7\text{x}+9+8]\Big]+\text{C}$
$=\frac{2}{49}\Big[(7\text{x}+9)^\frac{1}{2}[7\text{x}+17]\Big]+\text{C}$

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