What is the integral of 7/(xsqrt(4-49ln^2(x))) ?

The integral of 7/(xsqrt(4-49ln^2(x))) is arcsin(7/2 ln(x))+C

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integral of 7/(xsqrt(4-49ln^2(x)))

\int \frac{7}{x 4 - 49 ln ^{2} ( x )} d x

arcsin (\frac{7}{2} ln (x)) + C

Solution steps

\int \frac{7}{x 4 - 49 ln ^{2} ( x )} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{x 4 - 49 ln ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{4 - 49 u ^{2}} d u

= 7 \cdot \int \frac{1}{4 - 49 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{7} d v

= 7 \cdot \int \frac{1}{7} d v

Take the constant out:

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

= 7 \cdot \frac{1}{7} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 7 \cdot \frac{1}{7} \cdot 1 \cdot v

Substitute back

= 7 \cdot \frac{1}{7} \cdot 1 \cdot arcsin (\frac{7}{2} ln (x))

Simplify

7 \cdot \frac{1}{7} \cdot 1 \cdot arcsin (\frac{7}{2} ln (x)) : arcsin (\frac{7}{2} ln (x))

= arcsin (\frac{7}{2} ln (x))

Add a constant to the solution

= arcsin (\frac{7}{2} ln (x)) + C

\int 3 x (5 - x) d x

y^{^{''}} + 9 y = tan (3 x)

\int \frac{1}{12 x - 9 x ^{2}} d x

x \to 0 lim (\frac{x ^{2} + 3 x}{x})

\frac{( 1 + \frac{x}{2} )}{( 1 - \frac{x}{2} )}

What is the integral of 7/(xsqrt(4-49ln^2(x))) ?
The integral of 7/(xsqrt(4-49ln^2(x))) is arcsin(7/2 ln(x))+C