What is the integral of log_{10}(x^2+4x+5) ?

The integral of log_{10}(x^2+4x+5) is log_{10}(x^2+4x+5)(x+2)-2/(ln(10))(-arctan(x+2)+x+2)+C

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integral of log_{10}(x^2+4x+5)

integral

lo g_{10} (x^{2} + 4 x + 5)

lo g_{10} (x^{2} + 4 x + 5) (x + 2) - \frac{2}{ln ( 10 )} (- arctan (x + 2) + x + 2) + C

Solution steps

\int lo g_{10} (x^{2} + 4 x + 5) d x

Complete the square

x^{2} + 4 x + 5 : (x + 2)^{2} + 1

= \int lo g_{10} ((x + 2)^{2} + 1) d x

Apply u-substitution

: \int lo g_{10} (u^{2} + 1) d u

= \int lo g_{10} (u^{2} + 1) d u

Apply Integration By Parts

: u lo g_{10} (u^{2} + 1) - \int \frac{2 u ^{2}}{ln ( 10 ) ( u ^{2} + 1 )} d u

= u lo g_{10} (u^{2} + 1) - \int \frac{2 u ^{2}}{ln ( 10 ) ( u ^{2} + 1 )} d u

\int \frac{2 u ^{2}}{ln ( 10 ) ( u ^{2} + 1 )} d u = \frac{2}{ln ( 10 )} (- arctan (u) + u)

= u lo g_{10} (u^{2} + 1) - \frac{2}{ln ( 10 )} (- arctan (u) + u)

Substitute back

u = x + 2

= (x + 2) lo g_{10} ((x + 2)^{2} + 1) - \frac{2}{ln ( 10 )} (- arctan (x + 2) + x + 2)

Expand

(x + 2)^{2} + 1 : x^{2} + 4 x + 5

= lo g_{10} (x^{2} + 4 x + 5) (x + 2) - \frac{2}{ln ( 10 )} (- arctan (x + 2) + x + 2)

Add a constant to the solution

= lo g_{10} (x^{2} + 4 x + 5) (x + 2) - \frac{2}{ln ( 10 )} (- arctan (x + 2) + x + 2) + C

4 sec (x) tan (x) - 2 sec^{2} (x)

\frac{( 1 - 2 v - v ^{2} )}{( 1 + v )}

2 sin (x) - e^{x}

sin (x + 2 y)

(1 + x)^{0.5}

What is the integral of log_{10}(x^2+4x+5) ?
The integral of log_{10}(x^2+4x+5) is log_{10}(x^2+4x+5)(x+2)-2/(ln(10))(-arctan(x+2)+x+2)+C