What is the integral of 1/(sqrt(1+(7x+6)^2)) ?

The integral of 1/(sqrt(1+(7x+6)^2)) is 1/7 ln|7x+6+sqrt(49x^2+84x+37)|+C

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integral of 1/(sqrt(1+(7x+6)^2))

integral

\frac{1}{1 + ( 7 x + 6 ) ^{2}}

\frac{1}{7} ln 7 x + 6 + 49 x^{2} + 84 x + 37 + C

Solution steps

\int \frac{1}{1 + ( 7 x + 6 ) ^{2}} d x

Expand

1 + (7 x + 6)^{2} : 49 x^{2} + 84 x + 37

= \int \frac{1}{49 x ^{2} + 84 x + 37} d x

Complete the square

49 x^{2} + 84 x + 37 : 49 (x + \frac{6}{7})^{2} + 1

= \int \frac{1}{49 ( x + \frac{6}{7} ) ^{2} + 1} d x

Apply u-substitution

= \int \frac{1}{49 u ^{2} + 1} d u

Apply Trigonometric Substitution

= \int \frac{sec ( v )}{7} d v

Take the constant out:

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

= \frac{1}{7} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{7} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{7} ln tan (arctan (7 (x + \frac{6}{7}))) + sec (arctan (7 (x + \frac{6}{7})))

Simplify

\frac{1}{7} ln tan (arctan (7 (x + \frac{6}{7}))) + sec (arctan (7 (x + \frac{6}{7}))) : \frac{1}{7} ln 7 x + 6 + 49 x^{2} + 84 x + 37

= \frac{1}{7} ln 7 x + 6 + 49 x^{2} + 84 x + 37

Add a constant to the solution

= \frac{1}{7} ln 7 x + 6 + 49 x^{2} + 84 x + 37 + C

What is the integral of 1/(sqrt(1+(7x+6)^2)) ?
The integral of 1/(sqrt(1+(7x+6)^2)) is 1/7 ln|7x+6+sqrt(49x^2+84x+37)|+C