What is the integral of 7/((2x)^5) ?

The integral of 7/((2x)^5) is 7(1/(32x^4)-5/(128x^4))+C

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integral of 7/((2x)^5)

\int \frac{7}{( 2 x ) ^{5}} d x

7 (\frac{1}{32 x ^{4}} - \frac{5}{128 x ^{4}}) + C

Solution steps

\int \frac{7}{( 2 x ) ^{5}} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{( 2 x ) ^{5}} d x

Apply Integration By Parts

: \frac{1}{32 x ^{4}} - \int - \frac{5}{32 x ^{5}} d x

= 7 (\frac{1}{32 x ^{4}} - \int - \frac{5}{32 x ^{5}} d x)

\int - \frac{5}{32 x ^{5}} d x = \frac{5}{128 x ^{4}}

= 7 (\frac{1}{32 x ^{4}} - \frac{5}{128 x ^{4}})

Add a constant to the solution

= 7 (\frac{1}{32 x ^{4}} - \frac{5}{128 x ^{4}}) + C

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What is the integral of 7/((2x)^5) ?
The integral of 7/((2x)^5) is 7(1/(32x^4)-5/(128x^4))+C