What is the integral of 1/((2n-1)^2) ?

The integral of 1/((2n-1)^2) is -1/(2(2n-1))+C

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integral of 1/((2n-1)^2)

integral

\frac{1}{( 2 n - 1 ) ^{2}}

- \frac{1}{2 ( 2 n - 1 )} + C

Solution steps

\int \frac{1}{( 2 n - 1 ) ^{2}} d n

Apply u-substitution

: \int \frac{1}{2 u ^{2}} d u

= \int \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{2} (- \frac{1}{u})

Substitute back

u = 2 n - 1

= \frac{1}{2} (- \frac{1}{2 n - 1})

Simplify

\frac{1}{2} (- \frac{1}{2 n - 1}) : - \frac{1}{2 ( 2 n - 1 )}

= - \frac{1}{2 ( 2 n - 1 )}

Add a constant to the solution

= - \frac{1}{2 ( 2 n - 1 )} + C

\frac{2 x}{x + 2}

\frac{x ^{2}}{( x ^{2} + x + 1 ) ^{2}}

(sin (x) + cos^{2} (x))^{3}

\frac{2 ^{x}}{ln ( 2 )}

sin^{4} (x) + cos^{4} (x)

What is the integral of 1/((2n-1)^2) ?
The integral of 1/((2n-1)^2) is -1/(2(2n-1))+C