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

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

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

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

arctan (4 t^{2} - 1) + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (u)

Substitute back

u = 4 t^{2} - 1

= arctan (4 t^{2} - 1)

Add a constant to the solution

= arctan (4 t^{2} - 1) + C

y^{^{''}} - y = x^{2} e^{x}

\int 2 (x - 1) d x

y^{^{''}} + 2 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int_{1}^{3} (x - 1) d x

\frac{d}{d x} (2 tan (x) sec^{2} (x))

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