Question

Find the following integral. Sx V6x + 7 dx 5x 476x +7 dx = D.

Answer 1

$Given: \int x(6x+4)^{\frac{1}{4}}dx$

By using Integral by parts :

| f(x) + g(a)dır = [s(a) * [9(2) - / de f(a) * 912 g(2) dx

Here :

• 
  f(0) = =C
  
• 
  g(x) = (6x + 4) 4
  

EQUATIONS USED

• 
  ut ch+1 n+1 +0
  
• 
  constant * f (x)dt = constant * I f (2) da
  

$\Rightarrow \int x(6x+4)^{\frac{1}{4}}dx=\left [ x*\int (6x+4)^{\frac{1}{4}} \right ] -\int \left [ \frac{\mathrm{d} }{\mathrm{d} x}(x)*\int (6x+4)^{\frac{1}{4}} \right ]dx$

• $\int (6x+4)^{\frac{1}{4}}dx =\frac{(6x+4)^{\frac{1}{4}+1}}{\frac{1}{4}+1} \div \frac{\mathrm{d} }{\mathrm{d} x}(6x+4)$

  $=\frac{(6x+4)^{\frac{1+4}{4}}}{\frac{1+4}{4}} \div (6*1+0)$

$=\frac{(6x+4)^{\frac{5}{4}}}{\frac{5}{4}} \div 6$

  $=\frac{4}{5}*(6x+4)^{\frac{5}{4}} *\frac{1}{6}$

$=\frac{2}{5}*(6x+4)^{\frac{5}{4}} *\frac{1}{3}$

$=\frac{2}{15}(6x+4)^{\frac{5}{4}}$

$\Rightarrow \int x(6x+4)^{\frac{1}{4}}dx=\left [ x*\frac{2}{15}(6x+4)^{\frac{5}{4}} \right ] -\int \left [1*\frac{2}{15}(6x+4)^{\frac{5}{4}} \right ]dx$

  $=\left [ \frac{2x}{15}(6x+4)^{\frac{5}{4}} \right ] -\int \left [\frac{2}{15}(6x+4)^{\frac{5}{4}} \right ]dx$

  $=\left [ \frac{2x}{15}(6x+4)^{\frac{5}{4}} \right ] -\frac{2}{15}\int \left [(6x+4)^{\frac{5}{4}} \right ]dx$

• $\int (6x+4)^{\frac{5}{4}}dx =\frac{(6x+4)^{\frac{5}{4}+1}}{\frac{5}{4}+1} \div \frac{\mathrm{d} }{\mathrm{d} x}(6x+4)$

  $=\frac{(6x+4)^{\frac{5+4}{4}}}{\frac{5+4}{4}} \div (6*1+0)$

$=\frac{(6x+4)^{\frac{9}{4}}}{\frac{9}{4}} \div 6$

  $=\frac{4}{9}*(6x+4)^{\frac{9}{4}} *\frac{1}{6}$

$=\frac{2}{9}*(6x+4)^{\frac{9}{4}} *\frac{1}{3}$

$=\frac{2}{27}(6x+4)^{\frac{9}{4}}$

$\Rightarrow \int x(6x+4)^{\frac{1}{4}}dx=\left [ \frac{2x}{15}(6x+4)^{\frac{5}{4}} \right ] -\frac{2}{15}*\frac{2}{27}(6x+4)^{\frac{9}{4}}$

$\Rightarrow {\color{Red} \int x\sqrt[4]{(6x+4)}dx= \frac{2x}{15}(6x+4)^{\frac{5}{4}} -\frac{4}{405}(6x+4)^{\frac{9}{4}}}$