What is the integral of 1/(x^2+10x+41) ?

The integral of 1/(x^2+10x+41) is 1/4 arctan((x+5)/4)+C

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integral of 1/(x^2+10x+41)

\int \frac{1}{x ^{2} + 10 x + 41} d x

\frac{1}{4} arctan (\frac{x + 5}{4}) + C

Solution steps

\int \frac{1}{x ^{2} + 10 x + 41} d x

Complete the square

x^{2} + 10 x + 41 : (x + 5)^{2} + 16

= \int \frac{1}{( x + 5 ) ^{2} + 16} d x

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{x + 5}{4})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{x + 5}{4}) + C

(0, - 3), (- 3, 0)

\frac{d y}{d x} = \frac{( x + y ^{2} )}{2 y}

\frac{\partial}{\partial x} (x^{a} y^{b})

\frac{\partial}{\partial x} (\frac{( 2 x )}{y})

y^{^{''}} - 4 y = 6 e^{t}

What is the integral of 1/(x^2+10x+41) ?
The integral of 1/(x^2+10x+41) is 1/4 arctan((x+5)/4)+C