What is the integral of 1/(sqrt(x^2+6x+10)) ?

The integral of 1/(sqrt(x^2+6x+10)) is ln|sqrt(x^2+6x+10)+x+3|+C

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integral of 1/(sqrt(x^2+6x+10))

\int \frac{1}{x ^{2} + 6 x + 10} d x

ln x^{2} + 6 x + 10 + x + 3 + C

Solution steps

\int \frac{1}{x ^{2} + 6 x + 10} d x

Complete the square

x^{2} + 6 x + 10 : (x + 3)^{2} + 1

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

Apply u-substitution

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

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

Use the common integral:

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

= ln u^{2} + 1 + u

Substitute back

u = x + 3

= ln (x + 3)^{2} + 1 + x + 3

Expand

(x + 3)^{2} + 1 : x^{2} + 6 x + 10

= ln x^{2} + 6 x + 10 + x + 3

Add a constant to the solution

= ln x^{2} + 6 x + 10 + x + 3 + C

\int 33 sec^{4} (x) d x

y^{^{'}} = t^{2} y

y^{^{'}} = y + 2

\int \frac{( ( ln ( x ) ) ^{5} )}{x} d x

x ln (x) y^{^{'}} + y = 2 ln (x)

What is the integral of 1/(sqrt(x^2+6x+10)) ?
The integral of 1/(sqrt(x^2+6x+10)) is ln|sqrt(x^2+6x+10)+x+3|+C