Find the indefinite integral ∫...

MathematicsIntegration by SubstitutionJEE Advanced 1992Moderate

Find the indefinite integral $\int (\frac{1}{3 x + 4 x} + \frac{l n ( 1 + 6 x )}{3 x + x}) d x$

Answer:

\frac{3}{2} x^{2/3} - \frac{12}{7} x^{7/12} + 2 x^{1/2} - \frac{12}{5} x^{5/12} + 3 x^{1/3} - 4 x^{1/4} + 6 x^{1/6} - 12 x^{1/12} + 12 lo g ∣ x^{1/12} + 1∣ + 6 {\frac{( 1 + x ^{1/6} ) ^{3}}{3} - (1 + x^{1/6})^{2} + 3 (1 + x^{1/6})} ln (1 + x^{1/6}) - {\frac{( 1 + x ^{1/6} ) ^{3}}{9} - \frac{3}{4} (1 + x^{1/6})^{2} + 3 (1 + x^{1/6})} + \frac{( l n ( 1 + x ^{1/6} ) ) ^{2}}{2} + C

Visualized Solution (Hindi)

Splitting the Integral

Let $I = I_{1} + I_{2}$
$I_{1} = \int \frac{d x}{x ^{\frac{1}{3}} + x ^{\frac{1}{4}}}$
$I_{2} = \int \frac{l n ( 1 + x ^{\frac{1}{6}} )}{x ^{\frac{1}{3}} + x ^{\frac{1}{2}}} d x$

Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I_{1}

For $I_{1}$ , identify the fractional powers: $\frac{1}{3}$ and $\frac{1}{4}$
Find the LCM of denominators: $L C M (3, 4) = 12$
Substitute $x = y^{12} \Rightarrow d x = 12 y^{11} d y$

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I_{1}

$I_{1} = \int \frac{12 y ^{11}}{y ^{4} + y ^{3}} d y$
Factor out $y^{3}$ from the denominator: $I_{1} = 12 \int \frac{y ^{11}}{y ^{3} ( y + 1 )} d y$
Simplify the expression: $I_{1} = 12 \int \frac{y ^{8}}{y + 1} d y$

Polynomial Division

Perform polynomial division for $\frac{y ^{8}}{y + 1}$
$\frac{y ^{8}}{y + 1} = y^{7} - y^{6} + y^{5} - y^{4} + y^{3} - y^{2} + y - 1 + \frac{1}{y + 1}$

Integrating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I_{1}

$I_{1} = 12 [\frac{y ^{8}}{8} - \frac{y ^{7}}{7} + \frac{y ^{6}}{6} - \frac{y ^{5}}{5} + \frac{y ^{4}}{4} - \frac{y ^{3}}{3} + \frac{y ^{2}}{2} - y + ln ∣ y + 1∣]$
Substitute $y = x^{\frac{1}{12}}$ :
$I_{1} = \frac{3}{2} x^{\frac{2}{3}} - \frac{12}{7} x^{\frac{7}{12}} + 2 x^{\frac{1}{2}} - \frac{12}{5} x^{\frac{5}{12}} + 3 x^{\frac{1}{3}} - 4 x^{\frac{1}{4}} + 6 x^{\frac{1}{6}} - 12 x^{\frac{1}{12}} + 12 ln ∣ x^{\frac{1}{12}} + 1∣$

Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I_{2}

For $I_{2}$ , identify powers: $\frac{1}{6}, \frac{1}{3}, \frac{1}{2}$ . $L C M (6, 3, 2) = 6$
Substitute $x = z^{6} \Rightarrow d x = 6 z^{5} d z$
$I_{2} = \int \frac{l n ( 1 + z )}{z ^{2} + z ^{3}} 6 z^{5} d z = 6 \int \frac{z ^{3} l n ( 1 + z )}{1 + z} d z$

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I_{2} with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Substitute $1 + z = t \Rightarrow d z = d t$ and $z = t - 1$
$I_{2} = 6 \int \frac{( t - 1 ) ^{3} l n t}{t} d t$

Expanding the Term

Expand $(t - 1)^{3} = t^{3} - 3 t^{2} + 3 t - 1$
Divide by $t$ : $\frac{( t - 1 ) ^{3}}{t} = t^{2} - 3 t + 3 - \frac{1}{t}$
The integral becomes: $6 \int (t^{2} - 3 t + 3 - \frac{1}{t}) ln t d t$

Integration by Parts

Use Integration by Parts: $\int u d v = uv - \int v d u$ . Let $u = ln t$
$I_{2} = 6 [(\frac{t ^{3}}{3} - \frac{3 t ^{2}}{2} + 3 t) ln t - \int (\frac{t ^{2}}{3} - \frac{3 t}{2} + 3) d t - \int \frac{l n t}{t} d t]$
$I_{2} = 6 [(\frac{t ^{3}}{3} - \frac{3 t ^{2}}{2} + 3 t) ln t - (\frac{t ^{3}}{9} - \frac{3 t ^{2}}{4} + 3 t) - \frac{( l n t ) ^{2}}{2}]$

Final Result

Combine $I = I_{1} + I_{2}$
Substitute $t = 1 + x^{\frac{1}{6}}$ back into the $I_{2}$ result
Add the constant of integration $C$
Key Takeaway: Use LCM of fractional powers for radical substitutions and stay organized during long calculations.

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