What is the integral of 1/(t^2+6t+25) ?

The integral of 1/(t^2+6t+25) is 1/4 arctan((t+3)/4)+C

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integral of 1/(t^2+6t+25)

\int \frac{1}{t ^{2} + 6 t + 25} d t

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

Solution steps

\int \frac{1}{t ^{2} + 6 t + 25} d t

Complete the square

t^{2} + 6 t + 25 : (t + 3)^{2} + 16

= \int \frac{1}{( t + 3 ) ^{2} + 16} d t

Apply u-substitution

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

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

Apply Integral Substitution

= \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

Substitute back

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

Add a constant to the solution

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

(3, 2), (7, - 4)

x^{2}, 2 x - x^{2}, 0, 1

\int x (x^{4} - 1)^{2} d x

\frac{d}{d x} (x + ln (x^{2} - 1))

x \to - 2 π lim (sin (x))

What is the integral of 1/(t^2+6t+25) ?
The integral of 1/(t^2+6t+25) is 1/4 arctan((t+3)/4)+C