Evaluate. Assume u > O when In u appears. (In x)96 1 dex X O 96(In x)95+C (In x97 97x +C O (In x)97 +C O (In x)97 97 +
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EXPLANATION
[tex]\int \frac{(\ln x)^{96}}{x}dx[/tex]Applying subtitution: u=ln(x)
By integral substitution definition
[tex]\int f(g(x))\cdot g^{^{\prime}}(x)dx=\text{ }\int f(u)du,\text{ u=g(x)}[/tex]Substitute: u=ln(x)
[tex]\frac{du}{dx}=\frac{1}{x}[/tex][tex]\frac{d}{dx}=(\ln (x))[/tex]Apply the common derivative:
[tex]\frac{d}{dx}(\ln (x))=\frac{1}{x}[/tex][tex]\Rightarrow du=\frac{1}{x}dx[/tex][tex]\Rightarrow dx=xdu[/tex][tex]=\int \frac{u^{96}}{x}\text{xdu}[/tex]Simplify:
[tex]\frac{u^{96}}{x}x[/tex]Multiply fractions:
[tex]a\cdot\frac{b}{c}=\frac{a\cdot b}{c}[/tex][tex]=\frac{u^{96}x}{x}[/tex]Cancel the common factor: x
[tex]=u^{96}[/tex][tex]=\int u^{96}du[/tex]Apply the Power Rule:
[tex]\int x^adx=\frac{x^{(a+1)}}{a+1},\text{ a }\ne\text{ -1}[/tex][tex]=\frac{u^{96+1}}{96+1}[/tex]Substitute back u=ln(x)
[tex]=\frac{\ln ^{96+1}(x)}{96+1}[/tex]Simplify:
[tex]\frac{\ln ^{96+1}(x)}{96+1}[/tex]Add the numbers: 96+1=97
[tex]=\frac{\ln ^{97}(x)}{97}[/tex][tex]=\frac{1}{97}\ln ^{97}(x)[/tex]Add a constant to the solution:
[tex]=\frac{1}{97}\ln ^{97}(x)\text{ + C}[/tex]The answer is D:
[tex]\frac{(\ln x)^{97}}{97}+C[/tex]