What is the integral of 1/(sqrt(2(x-3)^2+1)) ?

The integral of 1/(sqrt(2(x-3)^2+1)) is 1/(sqrt(2))ln|sqrt(2)(x-3)+sqrt(2x^2-12x+19)|+C

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integral of 1/(sqrt(2(x-3)^2+1))

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

\frac{1}{2} ln 2 (x - 3) + 2 x^{2} - 12 x + 19 + C

Solution steps

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

Expand

2 (x - 3)^{2} + 1 : 2 x^{2} - 12 x + 19

= \int \frac{1}{2 x ^{2} - 12 x + 19} d x

Complete the square

2 x^{2} - 12 x + 19 : 2 (x - 3)^{2} + 1

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

Apply u-substitution

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

Apply Trigonometric Substitution

= \int \frac{sec ( v )}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{2} ln tan (arctan (2 (x - 3))) + sec (arctan (2 (x - 3)))

Simplify

\frac{1}{2} ln tan (arctan (2 (x - 3))) + sec (arctan (2 (x - 3))) : \frac{1}{2} ln 2 (x - 3) + 2 x^{2} - 12 x + 19

= \frac{1}{2} ln 2 (x - 3) + 2 x^{2} - 12 x + 19

Add a constant to the solution

= \frac{1}{2} ln 2 (x - 3) + 2 x^{2} - 12 x + 19 + C

\int x x + \frac{1}{x ^{3}} d x

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

\int 16 x tan (4 x^{2} + 1) d x

\int (- 10 + \frac{7}{p}) d p

What is the integral of 1/(sqrt(2(x-3)^2+1)) ?
The integral of 1/(sqrt(2(x-3)^2+1)) is 1/(sqrt(2))ln|sqrt(2)(x-3)+sqrt(2x^2-12x+19)|+C