What is the integral of (10x-15)/(x^2+25) ?

The integral of (10x-15)/(x^2+25) is 5ln|(x^2)/(25)+1|-3arctan(x/5)+C

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integral of (10x-15)/(x^2+25)

\int \frac{10 x - 15}{x ^{2} + 25} d x

5 ln \frac{x ^{2}}{25} + 1 - 3 arctan (\frac{x}{5}) + C

Solution steps

\int \frac{10 x - 15}{x ^{2} + 25} d x

Apply Integral Substitution

= \int \frac{10 u - 3}{u ^{2} + 1} d u

Expand

\frac{10 u - 3}{u ^{2} + 1} : \frac{10 u}{u ^{2} + 1} - \frac{3}{u ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{10 u}{u ^{2} + 1} d u - \int \frac{3}{u ^{2} + 1} d u

\int \frac{10 u}{u ^{2} + 1} d u = 5 ln u^{2} + 1

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

= 5 ln u^{2} + 1 - 3 arctan (u)

Substitute back

u = \frac{x}{5}

= 5 ln (\frac{x}{5})^{2} + 1 - 3 arctan (\frac{x}{5})

(\frac{x}{5})^{2} = \frac{x ^{2}}{25}

= 5 ln \frac{x ^{2}}{25} + 1 - 3 arctan (\frac{x}{5})

Add a constant to the solution

= 5 ln \frac{x ^{2}}{25} + 1 - 3 arctan (\frac{x}{5}) + C

\int_{0}^{\frac{π}{9}} x tan^{2} (3 x) d x

\frac{d}{d x} (\frac{ln ( sin ( x ^{2} ) + 1 )}{f ( x )})

y = x^{2} - 2, (5, 23)

x \to 1 + lim (x^{\frac{1}{1 - x}})

\int \frac{2 x}{1 - x ^{2}} d x

What is the integral of (10x-15)/(x^2+25) ?
The integral of (10x-15)/(x^2+25) is 5ln|(x^2)/(25)+1|-3arctan(x/5)+C