What is the integral of 1/((1-0.04x)^{0.3)} ?

The integral of 1/((1-0.04x)^{0.3)} is -35.71428…(1-0.04x)^{0.7}+C

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integral of 1/((1-0.04x)^{0.3)}

\int \frac{1}{( 1 - 0.04 x ) ^{0.3}} d x

- 35.71428 \dots (1 - 0.04 x)^{0.7} + C

Solution steps

\int \frac{1}{( 1 - 0.04 x ) ^{0.3}} d x

Apply u-substitution

: \int - \frac{1}{0.04 u ^{0.3}} d u

= \int - \frac{1}{0.04 u ^{0.3}} d u

Take the constant out:

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

= - \frac{1}{0.04} \cdot \int \frac{1}{u ^{0.3}} d u

Simplify

= - 25 \cdot \int \frac{1}{u ^{0.3}} d u

Apply the Power Rule

: 1.42857 \dots u^{0.7}

= - 25 \cdot 1.42857 \dots u^{0.7}

Substitute back

u = 1 - 0.04 x

= - 25 \cdot 1.42857 \dots (1 - 0.04 x)^{0.7}

Simplify

= - 35.71428 \dots (1 - 0.04 x)^{0.7}

Add a constant to the solution

= - 35.71428 \dots (1 - 0.04 x)^{0.7} + C

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What is the integral of 1/((1-0.04x)^{0.3)} ?
The integral of 1/((1-0.04x)^{0.3)} is -35.71428…(1-0.04x)^{0.7}+C