Question

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 +

Answer 1

EXPLANATION

`\int ((\ln x)^(96))/(x)dx`

Applying subtitution: u=ln(x)

By integral substitution definition

`\int f(g(x))\cdot g^{^(\prime)}(x)dx=\text{ }\int f(u)du,\text{ u=g(x)}`

Substitute: u=ln(x)

`(du)/(dx)=(1)/(x)`
`(d)/(dx)=(\ln (x))`

Apply the common derivative:

`(d)/(dx)(\ln (x))=(1)/(x)`
`\Rightarrow du=(1)/(x)dx`
`\Rightarrow dx=xdu`
`=\int (u^(96))/(x)\text{xdu}`

Simplify:

`(u^(96))/(x)x`

Multiply fractions:

`a\cdot(b)/(c)=(a\cdot b)/(c)`
`=(u^(96)x)/(x)`

Cancel the common factor: x

`=u^(96)`
`=\int u^(96)du`

Apply the Power Rule:

`\int x^adx=(x^((a+1)))/(a+1),\text{ a }\\e\text{ -1}`
`=(u^(96+1))/(96+1)`

Substitute back u=ln(x)

`=(\ln ^(96+1)(x))/(96+1)`

Simplify:

`(\ln ^(96+1)(x))/(96+1)`

Add the numbers: 96+1=97

`=(\ln ^(97)(x))/(97)`
`=(1)/(97)\ln ^(97)(x)`

Add a constant to the solution:

`=(1)/(97)\ln ^(97)(x)\text{ + C}`

The answer is D:

`((\ln x)^(97))/(97)+C`