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

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

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

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

\frac{1}{2} arcsec (\frac{1}{4} (t + 3)) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Apply Trigonometric Substitution

= 2 \cdot \int \frac{1}{4} d v

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{4} \cdot 1 \cdot v

Substitute back

= 2 \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{1}{4} (t + 3))

Simplify

2 \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{1}{4} (t + 3)) : \frac{1}{2} arcsec (\frac{1}{4} (t + 3))

= \frac{1}{2} arcsec (\frac{1}{4} (t + 3))

Add a constant to the solution

= \frac{1}{2} arcsec (\frac{1}{4} (t + 3)) + C

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