What is the integral of 1/(t^2+6t+50) ?

The integral of 1/(t^2+6t+50) is 1/(sqrt(41))arctan((t+3)/(sqrt(41)))+C

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integral of 1/(t^2+6t+50)

\int \frac{1}{t ^{2} + 6 t + 50} d t

\frac{1}{41} arctan (\frac{t + 3}{41}) + C

Solution steps

\int \frac{1}{t ^{2} + 6 t + 50} d t

Complete the square

t^{2} + 6 t + 50 : (t + 3)^{2} + 41

= \int \frac{1}{( t + 3 ) ^{2} + 41} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 41} d u

= \int \frac{1}{u ^{2} + 41} d u

Apply Integral Substitution

= \int \frac{1}{41 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{41} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{41} arctan (v)

Substitute back

= \frac{1}{41} arctan (\frac{t + 3}{41})

Add a constant to the solution

= \frac{1}{41} arctan (\frac{t + 3}{41}) + C

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What is the integral of 1/(t^2+6t+50) ?
The integral of 1/(t^2+6t+50) is 1/(sqrt(41))arctan((t+3)/(sqrt(41)))+C