What is the integral of (2xln(x))/(-x^4) ?

The integral of (2xln(x))/(-x^4) is (ln(x))/(x^2)+1/(2x^2)+C

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integral of (2xln(x))/(-x^4)

\int \frac{2 x ln ( x )}{- x ^{4}} d x

\frac{ln ( x )}{x ^{2}} + \frac{1}{2 x ^{2}} + C

Solution steps

\int \frac{2 x ln ( x )}{- x ^{4}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{- 1} \cdot \int \frac{x ln ( x )}{x ^{4}} d x

Simplify

= (- 2) \cdot \int \frac{x ln ( x )}{x ^{4}} d x

Cancel the common factor:

x

= (- 2) \cdot \int \frac{ln ( x )}{x ^{3}} d x

Apply Integration By Parts

: - \frac{ln ( x )}{2 x ^{2}} - \int - \frac{1}{2 x ^{3}} d x

= (- 2) (- \frac{ln ( x )}{2 x ^{2}} - \int - \frac{1}{2 x ^{3}} d x)

\int - \frac{1}{2 x ^{3}} d x = \frac{1}{4 x ^{2}}

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

Simplify

(- 2) (- \frac{ln ( x )}{2 x ^{2}} - \frac{1}{4 x ^{2}}) : \frac{ln ( x )}{x ^{2}} + \frac{1}{2 x ^{2}}

= \frac{ln ( x )}{x ^{2}} + \frac{1}{2 x ^{2}}

Add a constant to the solution

= \frac{ln ( x )}{x ^{2}} + \frac{1}{2 x ^{2}} + C

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y^{^{''}} + 9 y = tan (3 x)

What is the integral of (2xln(x))/(-x^4) ?
The integral of (2xln(x))/(-x^4) is (ln(x))/(x^2)+1/(2x^2)+C