What is the integral of 1/((x+3)sqrt(x^2+6x-7)) ?

The integral of 1/((x+3)sqrt(x^2+6x-7)) is 1/4 arcsec(1/4 (x+3))+C

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integral of 1/((x+3)sqrt(x^2+6x-7))

\int \frac{1}{( x + 3 ) x ^{2} + 6 x - 7} d x

\frac{1}{4} arcsec (\frac{1}{4} (x + 3)) + C

Solution steps

\int \frac{1}{( x + 3 ) x ^{2} + 6 x - 7} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 16} d u

= \int \frac{1}{u u ^{2} - 16} d u

Apply Trigonometric Substitution

: \int \frac{1}{4} d v

= \int \frac{1}{4} d v

Take the constant out:

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

= \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{4} \cdot 1 \cdot v

Substitute back

= \frac{1}{4} \cdot 1 \cdot arcsec (\frac{1}{4} (x + 3))

Simplify

= \frac{1}{4} arcsec (\frac{1}{4} (x + 3))

Add a constant to the solution

= \frac{1}{4} arcsec (\frac{1}{4} (x + 3)) + C

\int 0.0015 x^{2} d x

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

x \to 2 lim (x^{2} + 3 x + 1)

x \to 0 lim (\frac{ln ( sin ( 2 x ) )}{ln ( tan ( x ) )})

\int \frac{1}{9 - 81 x ^{2}} d x

What is the integral of 1/((x+3)sqrt(x^2+6x-7)) ?
The integral of 1/((x+3)sqrt(x^2+6x-7)) is 1/4 arcsec(1/4 (x+3))+C