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Popular >

integral of 1/(2x-9)

Solution

\int \frac{1}{2 x - 9} d x

Solution

\frac{1}{2} ln ∣ 2 x - 9 ∣ + C

Solution steps

\int \frac{1}{2 x - 9} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 9

= \frac{1}{2} ln ∣ 2 x - 9 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x - 9 ∣ + C

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\int (e^{\frac{x}{2}})^{2} d x

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\int cos^{17} (x) sin (x) d x

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\int (9 x^{2} - 4 x + 3) d x

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

integral of x(x+4)^4

\int x (x + 4)^{4} d x

Frequently Asked Questions (FAQ)

What is the integral of 1/(2x-9) ?
The integral of 1/(2x-9) is 1/2 ln|2x-9|+C