What is the integral of (3t)/(sqrt(1-9t^4)) ?

The integral of (3t)/(sqrt(1-9t^4)) is 1/2 arcsin(3t^2)+C

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integral of (3t)/(sqrt(1-9t^4))

\int \frac{3 t}{1 - 9 t ^{4}} d t

\frac{1}{2} arcsin (3 t^{2}) + C

Solution steps

\int \frac{3 t}{1 - 9 t ^{4}} d t

Take the constant out:

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

= 3 \cdot \int \frac{t}{1 - 9 t ^{4}} d t

Apply u-substitution

: \int \frac{1}{2 1 - 9 u ^{2}} d u

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{3} d v

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 3 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot v

Substitute back

= 3 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot arcsin (3 t^{2})

Simplify

3 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot arcsin (3 t^{2}) : \frac{1}{2} arcsin (3 t^{2})

= \frac{1}{2} arcsin (3 t^{2})

Add a constant to the solution

= \frac{1}{2} arcsin (3 t^{2}) + C

\int 3 cos (3 x) d x

f (x) = 5 e^{x} + 7 x, at x = 0

f (x) = - 3 x^{2} - 5 x + 1, at x = 1

\frac{\partial}{\partial t} (e^{2 x} - x cos (t))

f (x) = x^{2} + 8 x - 2

What is the integral of (3t)/(sqrt(1-9t^4)) ?
The integral of (3t)/(sqrt(1-9t^4)) is 1/2 arcsin(3t^2)+C