# How do I evaluate the following indefinite integral by using integration by parts? int(7lnx)/x^3dx

## $\int \frac{7 \ln x}{x} ^ 3 \mathrm{dx}$

Jan 19, 2018

$\int \frac{7 \ln x}{x} ^ 3 \mathrm{dx} = - \frac{7 \ln x}{2 {x}^{2}} - \frac{7}{4 {x}^{2}} + \text{C}$

#### Explanation:

Given: $\int \frac{7 \ln x}{x} ^ 3 \mathrm{dx}$

Rewrite the integral as:

$7 \int \ln \frac{x}{x} ^ 3 \implies 7 \int \ln x \cdot \frac{1}{x} ^ 3 \mathrm{dx}$

Apply Integration by Parts (I.B.P)

$\int$ $u$ $\mathrm{dv} = u v - \int$ $v$ $\mathrm{du}$

Let:
$u = \ln x \implies \mathrm{du} = \frac{1}{x} \mathrm{dx}$

$\mathrm{dv} = \frac{1}{x} ^ 3 \mathrm{dx} \implies v = - \frac{1}{2 {x}^{2}} \leftarrow$ See proof below

$- - - - - - - - - - - - - - - - - - - -$

Use color(blue)(intx^adx=(x^(a+1))/(a+1)

color(red)(int1/x^3dx=intx^(-3)dx=(x^(-3+1))/(-3+1)=x^-2/-2=-1/(2x^2)

$- - - - - - - - - - - - - - - - - - - -$

Substituting into the I.P.B formula:

$= \left(\ln x\right) \left(- \frac{1}{2 {x}^{2}}\right) - \int \left(- \frac{1}{2 {x}^{2}}\right) \left(\frac{1}{x} \mathrm{dx}\right)$

Take out the constant $- \frac{1}{2}$

$= - \ln \frac{x}{2 {x}^{2}} - - \frac{1}{2} \int \frac{1}{{x}^{3}} \mathrm{dx}$

Simplify

$= - \ln \frac{x}{2 {x}^{2}} + \frac{1}{2} \int \frac{1}{{x}^{3}}$

Now we solve for $\int \frac{1}{x} ^ 3$ using which we have already done earlier:

$= - \ln \frac{x}{2 {x}^{2}} + \frac{1}{2} \left[- \frac{1}{2 {x}^{2}}\right]$

$= - \ln \frac{x}{2 {x}^{2}} - \frac{1}{4 {x}^{2}}$

$= 7 \left[- \ln \frac{x}{2 {x}^{2}} - \frac{1}{4 {x}^{2}}\right]$

$= - \frac{7 \ln x}{2 {x}^{2}} - \frac{7}{4 {x}^{2}} + \text{C} \leftarrow$ Don't forget the constant!