What is the integral of 7/(xsqrt(36-49ln^2(x))) ?

The integral of 7/(xsqrt(36-49ln^2(x))) is arcsin(7/6 ln(x))+C

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integral of 7/(xsqrt(36-49ln^2(x)))

\int \frac{7}{x 36 - 49 ln ^{2} ( x )} d x

arcsin (\frac{7}{6} ln (x)) + C

Solution steps

\int \frac{7}{x 36 - 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 36 - 49 ln ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{36 - 49 u ^{2}} d u

= 7 \cdot \int \frac{1}{36 - 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}{6} ln (x))

Simplify

7 \cdot \frac{1}{7} \cdot 1 \cdot arcsin (\frac{7}{6} ln (x)) : arcsin (\frac{7}{6} ln (x))

= arcsin (\frac{7}{6} ln (x))

Add a constant to the solution

= arcsin (\frac{7}{6} ln (x)) + C

\frac{d}{d x} (x sin (x) + cos (x))

\int 1 - cos (t) d t

\int 13 + 12 x - x^{2} d x

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

y^{^{'}} = 2 y + x^{2} + 7

What is the integral of 7/(xsqrt(36-49ln^2(x))) ?
The integral of 7/(xsqrt(36-49ln^2(x))) is arcsin(7/6 ln(x))+C