What is the integral of 1/((2x-3)^4) ?

The integral of 1/((2x-3)^4) is -1/(6(2x-3)^3)+C

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integral of 1/((2x-3)^4)

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

- \frac{1}{6 ( 2 x - 3 ) ^{3}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ^{4}} d u

= \int \frac{1}{2 u ^{4}} 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 ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{2} (- \frac{1}{3 u ^{3}})

Substitute back

u = 2 x - 3

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

Simplify

\frac{1}{2} (- \frac{1}{3 ( 2 x - 3 ) ^{3}}) : - \frac{1}{6 ( 2 x - 3 ) ^{3}}

= - \frac{1}{6 ( 2 x - 3 ) ^{3}}

Add a constant to the solution

= - \frac{1}{6 ( 2 x - 3 ) ^{3}} + C

\int_{1}^{\infty} x^{- \frac{1}{5}} d x

x \to 1 + lim (\frac{x + 1}{x ^{2} - 1})

\int 2 sin (x) + 3 cos (x) d x

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 13

\int_{0}^{1} x (x - x^{2}) d x

What is the integral of 1/((2x-3)^4) ?
The integral of 1/((2x-3)^4) is -1/(6(2x-3)^3)+C