What is the integral of 3/(4(4-3x)^7) ?

The integral of 3/(4(4-3x)^7) is 1/(24(3x-4)^6)+C

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integral of 3/(4(4-3x)^7)

\int \frac{3}{4 ( 4 - 3 x ) ^{7}} d x

\frac{1}{24 ( 3 x - 4 ) ^{6}} + C

Solution steps

\int \frac{3}{4 ( 4 - 3 x ) ^{7}} d x

Take the constant out:

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

= \frac{3}{4} \cdot \int \frac{1}{( 4 - 3 x ) ^{7}} d x

Take the partial fraction of

\frac{1}{( 4 - 3 x ) ^{7}} : - \frac{1}{( 3 x - 4 ) ^{7}}

= \frac{3}{4} \cdot \int - \frac{1}{( 3 x - 4 ) ^{7}} d x

Take the constant out:

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

= \frac{3}{4} (- \int \frac{1}{( 3 x - 4 ) ^{7}} d x)

Apply u-substitution

: \int \frac{1}{3 u ^{7}} d u

= \frac{3}{4} (- \int \frac{1}{3 u ^{7}} d u)

Take the constant out:

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

= \frac{3}{4} (- \frac{1}{3} \cdot \int \frac{1}{u ^{7}} d u)

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

= \frac{3}{4} (- \frac{1}{3} (- \frac{1}{6 u ^{6}}))

Substitute back

u = 3 x - 4

= \frac{3}{4} (- \frac{1}{3} (- \frac{1}{6 ( 3 x - 4 ) ^{6}}))

Simplify

\frac{3}{4} (- \frac{1}{3} (- \frac{1}{6 ( 3 x - 4 ) ^{6}})) : \frac{1}{24 ( 3 x - 4 ) ^{6}}

= \frac{1}{24 ( 3 x - 4 ) ^{6}}

Add a constant to the solution

= \frac{1}{24 ( 3 x - 4 ) ^{6}} + C

4 y^{^{''}} + 9 y = 0

\int \frac{4 x ^{3}}{x ^{4} + 7} d x

f (x) = 2 x^{2} + 5 x

\int 2 t d t

- x^{2} + 4 x + 1, x + 1

What is the integral of 3/(4(4-3x)^7) ?
The integral of 3/(4(4-3x)^7) is 1/(24(3x-4)^6)+C