# Find the most general antiderivative or indefinite integral. ? 3 + 4 x 1 + x 2 d x

## Question:

Find the most general antiderivative or indefinite integral.

$$\int \frac{3+4x}{1+x^2}\enspace dx$$

## Indefinite Integrals:

Indefinite Integrals really are the same as antiderivatives. Finding the most general one, means we want to leave the constant of integration as an unknown that can be adjusted to fit any particular conditions. To solve this one we will break it up.

Let's split the integral up here. We start by writing

{eq}\begin{align*} \int \frac{3+4x}{1+x^2}\ dx &= \int \frac3{1+x^2} \ dx + \int \frac{4x}{1+x^2}\ dx \end{align*} {/eq}

Now, we can use the fact that

{eq}\begin{align*} \frac{d}{dx} \tan^{-1} x = \frac1{1+x^2} \end{align*} {/eq}

and the fact that

{eq}\begin{align*} \frac{d}{dx} \ln (1+x^2) &= \frac{2x}{1+x^2} \end{align*} {/eq}

to finish things off. Note that the derivatives above are hiding in our equation, and that recognizing the second integrand as the derivative of a natural log function enables us to skip the substitution method by seeing that method for what it really is: using the fundamental theorem of calculus along with the chain rule in reverse. We rewrite again to take explicit advantage of the fundamental theorem of calculus and find

{eq}\begin{align*} \int \frac{3+4x}{1+x^2}\ dx &= \int \frac3{1+x^2} \ dx + \int \frac{4x}{1+x^2}\ dx \\ &= 3 \int \frac1{1+x^2} \ dx + 2 \int \frac{2x}{1+x^2}\ dx \\ &= 3 \left( \tan^{-1} x + C \right) + 2 \left( \ln(1+x^2) + D \right) \\ &= 3 \tan^{-1} x + 2 \ln (1+x^2) + (3C + 2D) \\ &= 3 \tan^{-1} x + 2 \ln (1+x^2) + K \end{align*} {/eq}

where, since constants are just numbers, we combined ours into one and wrote it as {eq}K = 3C+2D {/eq}.