What is the integral of 1/((k-1)^4) ?

The integral of 1/((k-1)^4) is x/((k-1)^4)+C

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integral of 1/((k-1)^4)

\int \frac{1}{( k - 1 ) ^{4}}

\frac{x}{( k - 1 ) ^{4}} + C

Solution steps

\int \frac{1}{( k - 1 ) ^{4}} d x

Integral of a constant:

\int a d x = a x

= \frac{1}{( k - 1 ) ^{4}} x

Simplify

\frac{1}{( k - 1 ) ^{4}} x : \frac{x}{( k - 1 ) ^{4}}

= \frac{x}{( k - 1 ) ^{4}}

Add a constant to the solution

= \frac{x}{( k - 1 ) ^{4}} + C

y = \frac{3 x ^{2} + 2}{x + 1}

\int \frac{1}{x + 1 + x} d x

\frac{d ^{3}}{d x ^{3}} (4 x - 3)

\int \frac{2 x + 6}{x ^{2} + 1} d x

x \to 2 - lim (3 - x)

What is the integral of 1/((k-1)^4) ?
The integral of 1/((k-1)^4) is x/((k-1)^4)+C