What is the integral of 2/((2t-1)^2+4) ?

The integral of 2/((2t-1)^2+4) is 1/2 arctan(1/2 (2t-1))+C

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integral of 2/((2t-1)^2+4)

\int \frac{2}{( 2 t - 1 ) ^{2} + 4} d t

\frac{1}{2} arctan (\frac{1}{2} (2 t - 1)) + C

Solution steps

\int \frac{2}{( 2 t - 1 ) ^{2} + 4} d t

Take the constant out:

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

= 2 \cdot \int \frac{1}{( 2 t - 1 ) ^{2} + 4} d t

Apply u-substitution

= 2 \cdot \int \frac{1}{2 ( u ^{2} + 4 )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} arctan (v)

Substitute back

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{2 t - 1}{2})

Simplify

2 \cdot \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{2 t - 1}{2}) : \frac{1}{2} arctan (\frac{1}{2} (2 t - 1))

= \frac{1}{2} arctan (\frac{1}{2} (2 t - 1))

Add a constant to the solution

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

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What is the integral of 2/((2t-1)^2+4) ?
The integral of 2/((2t-1)^2+4) is 1/2 arctan(1/2 (2t-1))+C