What is the integral of 1/(x^2sqrt(3x-1)) ?

The integral of 1/(x^2sqrt(3x-1)) is 3(arctan(sqrt(3x-1))+1/2 sin(2arctan(sqrt(3x-1))))+C

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integral of 1/(x^2sqrt(3x-1))

\int \frac{1}{x ^{2} 3 x - 1} d x

3 (arctan (3 x - 1) + \frac{1}{2} sin (2 arctan (3 x - 1))) + C

Solution steps

\int \frac{1}{x ^{2} 3 x - 1} d x

Apply u-substitution

: \int \frac{6}{( u ^{2} + 1 ) ^{2}} d u

= \int \frac{6}{( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= 6 \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= 6 \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= 6 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

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

= 6 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 6 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 6 \cdot \frac{1}{2} (arctan (3 x - 1) + \frac{1}{2} sin (2 arctan (3 x - 1)))

Simplify

6 \cdot \frac{1}{2} (arctan (3 x - 1) + \frac{1}{2} sin (2 arctan (3 x - 1))) : 3 (arctan (3 x - 1) + \frac{1}{2} sin (2 arctan (3 x - 1)))

= 3 (arctan (3 x - 1) + \frac{1}{2} sin (2 arctan (3 x - 1)))

Add a constant to the solution

= 3 (arctan (3 x - 1) + \frac{1}{2} sin (2 arctan (3 x - 1))) + C

\int 2 x (x^{2} + 3)^{4} d x

\frac{\partial}{\partial y} (y sin (x y))

\frac{d}{d x} ((ln (x))^{5})

\int 8 x e^{8 x} d x

ln^{3} (x)

What is the integral of 1/(x^2sqrt(3x-1)) ?
The integral of 1/(x^2sqrt(3x-1)) is 3(arctan(sqrt(3x-1))+1/2 sin(2arctan(sqrt(3x-1))))+C