What is the integral of (9e^{-9/8 x})/(32)x ?

The integral of (9e^{-9/8 x})/(32)x is -1/36 e^{-(9x)/8}(9x+8)+C

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integral of (9e^{-9/8 x})/(32)x

\int \frac{9 e ^{- \frac{9}{8} x}}{32} x d x

- \frac{1}{36} e^{- \frac{9 x}{8}} (9 x + 8) + C

Solution steps

\int \frac{9 e ^{- \frac{9}{8} x}}{32} x d x

Take the constant out:

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

= \frac{9}{32} \cdot \int e^{- \frac{9}{8} x} x d x

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{9}{32} \cdot \int e^{- \frac{9 x}{8}} x d x

Apply Integration By Parts

: - \frac{8}{9} e^{- \frac{9 x}{8}} x - \int - \frac{8}{9} e^{- \frac{9 x}{8}} d x

= \frac{9}{32} (- \frac{8}{9} e^{- \frac{9 x}{8}} x - \int - \frac{8}{9} e^{- \frac{9 x}{8}} d x)

\int - \frac{8}{9} e^{- \frac{9 x}{8}} d x = \frac{64}{81} e^{- \frac{9 x}{8}}

= \frac{9}{32} (- \frac{8}{9} e^{- \frac{9 x}{8}} x - \frac{64}{81} e^{- \frac{9 x}{8}})

Simplify

\frac{9}{32} (- \frac{8}{9} e^{- \frac{9 x}{8}} x - \frac{64}{81} e^{- \frac{9 x}{8}}) : - \frac{1}{36} e^{- \frac{9 x}{8}} (9 x + 8)

= - \frac{1}{36} e^{- \frac{9 x}{8}} (9 x + 8)

Add a constant to the solution

= - \frac{1}{36} e^{- \frac{9 x}{8}} (9 x + 8) + C

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What is the integral of (9e^{-9/8 x})/(32)x ?
The integral of (9e^{-9/8 x})/(32)x is -1/36 e^{-(9x)/8}(9x+8)+C