What is the integral of ((5t-t))/(sqrt(3t^2-9)) ?

The integral of ((5t-t))/(sqrt(3t^2-9)) is 4/3 sqrt(3t^2-9)+C

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integral of ((5t-t))/(sqrt(3t^2-9))

\int \frac{( 5 t - t )}{3 t ^{2} - 9} d t

\frac{4}{3} 3 t^{2} - 9 + C

Solution steps

\int \frac{5 t - t}{3 t ^{2} - 9} d t

Expand

\frac{5 t - t}{3 t ^{2} - 9} : \frac{5 t}{3 t ^{2} - 9} - \frac{t}{3 t ^{2} - 9}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{5 t}{3 t ^{2} - 9} d t - \int \frac{t}{3 t ^{2} - 9} d t

\int \frac{5 t}{3 t ^{2} - 9} d t = \frac{5}{3} 3 t^{2} - 9

\int \frac{t}{3 t ^{2} - 9} d t = \frac{1}{3} 3 t^{2} - 9

= \frac{5}{3} 3 t^{2} - 9 - \frac{1}{3} 3 t^{2} - 9

Simplify

\frac{5}{3} 3 t^{2} - 9 - \frac{1}{3} 3 t^{2} - 9 : \frac{4}{3} 3 t^{2} - 9

= \frac{4}{3} 3 t^{2} - 9

Add a constant to the solution

= \frac{4}{3} 3 t^{2} - 9 + C

\frac{\partial}{\partial x} (x^{4} - 3 x^{2} y^{2} + y^{4})

x \to - 2 lim (4^{2} - 3)

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = 0, y (\frac{π}{2}) = 1

f (x) = 4 x^{2} - 1

y = (1 + 5 x)^{12}, (0, 1)

What is the integral of ((5t-t))/(sqrt(3t^2-9)) ?
The integral of ((5t-t))/(sqrt(3t^2-9)) is 4/3 sqrt(3t^2-9)+C