What is the integral of 1/((x-43)^{0.98)} ?

The integral of 1/((x-43)^{0.98)} is 50(x-43)^{0.02}+C

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integral of 1/((x-43)^{0.98)}

\int \frac{1}{( x - 43 ) ^{0.98}} d x

50 (x - 43)^{0.02} + C

Solution steps

\int \frac{1}{( x - 43 ) ^{0.98}} d x

Apply u-substitution

: \int \frac{1}{u ^{0.98}} d u

= \int \frac{1}{u ^{0.98}} d u

Apply the Power Rule

: 50 u^{0.02}

= 50 u^{0.02}

Substitute back

u = x - 43

= 50 (x - 43)^{0.02}

Add a constant to the solution

= 50 (x - 43)^{0.02} + C

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\frac{\partial}{\partial x} (yz e^{x z})

y^{^{'''}} + y^{^{''}} + 8 y^{^{'}} - 10 y = 0

\frac{\partial}{\partial x} (\frac{1}{( x - 2 ) ^{2}})

What is the integral of 1/((x-43)^{0.98)} ?
The integral of 1/((x-43)^{0.98)} is 50(x-43)^{0.02}+C