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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
$\int\frac{\text{x}^2}{\sqrt{3\text{x}-4}}\text{dx}$
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Answer

$\int\frac{\text{x}^2}{\sqrt{3\text{x}-4}}\text{dx}$
Let $3\text{x}+4=\text{t}$
$\Rightarrow\text{x}=\frac{\text{t}^{-4}}{3}$
$\Rightarrow1=\frac{1}{3}\cdot\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{3}$
Now, $\int\frac{\text{x}^2}{\sqrt{3\text{x}-4}}$
$=\frac{1}{3}\int\frac{\Big(\frac{\text{t}^{-4}}{3}\Big)^2}{\sqrt{\text{t}}}\text{dt}$
$=\frac{1}{27}\int\Big(\frac{\text{t}^2}{\sqrt{\text{t}}}-\frac{8\text{t}}{\sqrt{\text{t}}}+\frac{16}{\sqrt{t}}\Big)\text{dt}$
$=\frac{1}{27}\int\Big(\text{t}^\frac{3}{2}-8\text{t}^\frac{1}{2}+16\text{t}^{-\frac{1}{2}}\Big)\text{dt}$
$=\frac{1}{27}\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{8\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{16\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=\frac{1}{27}\Big[\frac{2}{5}\text{t}^{\frac{5}{2}}-\frac{8\times2}{3}\text{t}^{\frac{3}{2}}+32\text{t}^{\frac{1}{2}}\Big]+\text{C}$
$=\frac{2}{135}(\text{t})^{\frac{5}{2}}-\frac{16}{81}\text{t}^{\frac{3}{2}}+\frac{32}{27}\text{t}^\frac{1}{2}+\text{C}$
$=\frac{2}{135}(3\text{x}+4)^{\frac{5}{2}}-\frac{16}{81}(3\text{x}+4)^{\frac{3}{2}}+\frac{32}{27}(3\text{x}+4)^{\frac{1}{2}}+\text{C}$

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