What is the integral of x/(sqrt(7x^2-3)) ?

The integral of x/(sqrt(7x^2-3)) is 1/7 sqrt(7x^2-3)+C

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integral of x/(sqrt(7x^2-3))

\int \frac{x}{7 x ^{2} - 3} d x

\frac{1}{7} 7 x^{2} - 3 + C

Solution steps

\int \frac{x}{7 x ^{2} - 3} d x

Apply u-substitution

: \int \frac{1}{14 u} d u

= \int \frac{1}{14 u} d u

Take the constant out:

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

= \frac{1}{14} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{14} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 7 x^{2} - 3

= \frac{1}{14} \cdot 2 (7 x^{2} - 3)^{\frac{1}{2}}

Simplify

\frac{1}{14} \cdot 2 (7 x^{2} - 3)^{\frac{1}{2}} : \frac{1}{7} 7 x^{2} - 3

= \frac{1}{7} 7 x^{2} - 3

Add a constant to the solution

= \frac{1}{7} 7 x^{2} - 3 + C

\int (sin^{2} (x)) d x

x \to a lim (f (a))

\int 9 x cos (4 x) d x

x = - 2, x = 2, y = x, y = x^{2} - 2

y^{^{''}} + a y - b = 0

What is the integral of x/(sqrt(7x^2-3)) ?
The integral of x/(sqrt(7x^2-3)) is 1/7 sqrt(7x^2-3)+C