What is the integral of 1/((x-43)^{0.92)} ?

The integral of 1/((x-43)^{0.92)} is 12.5(x-43)^{0.08}+C

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integral of 1/((x-43)^{0.92)}

\int \frac{1}{( x - 43 ) ^{0.92}} d x

12.5 (x - 43)^{0.08} + C

Solution steps

\int \frac{1}{( x - 43 ) ^{0.92}} d x

Apply u-substitution

: \int \frac{1}{u ^{0.92}} d u

= \int \frac{1}{u ^{0.92}} d u

Apply the Power Rule

: 12.5 u^{0.08}

= 12.5 u^{0.08}

Substitute back

u = x - 43

= 12.5 (x - 43)^{0.08}

Add a constant to the solution

= 12.5 (x - 43)^{0.08} + C

co tan (x)

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

\int (8 x + 7) d x

f (x) = \frac{1 - x}{1 + x}

f (x) = x^{2} + 15, at x = 7

What is the integral of 1/((x-43)^{0.92)} ?
The integral of 1/((x-43)^{0.92)} is 12.5(x-43)^{0.08}+C