Find int(1)/(4x^(2)+48 x+148)dx. Choose 1 answer: (A) (1)/(4)arcsin(x+6)+C (B) (1)/(4)arcsin((x+6)/(4))+C (C) (1)/(4)arctan(x+6)+C (D) (1)/(4)arctan((x+6)/(4))+C

Complete the Square: First, complete the square for the quadratic in the denominator. 4x^2 + 48x + 148 can be written as (2x)^2 + 2*2x*6 + 6^2 + (148 - 36). This simplifies to (2x + 6)^2 + 112.Rewrite with Completed Square: Now, rewrite the integral with the completed square. int(1)/(4x^2 + 48x + 148)dx = int(1)/((2x + 6)^2 + 112)dx.Factor Out 4: Factor out the 4 from the denominator to match the form of the arctan integral. int(1)/((2x + 6)^2 + 112)dx = int(1)/(4*((x + 3)^2 + 28))dx.Recognize Integral Form: Recognize that the integral is now in the form of int(1)/(a^2 + u^2)du, which is 1/a * arctan(u/a) + C. Here, a^2 = 28 and u = x + 3, so a = sqrt(28) = 2sqrt(7).Calculate Integral: Calculate the integral using the arctan formula. int(1)/(4*((x + 3)^2 + 28))dx = 1/(4*2sqrt(7)) * arctan((x + 3)/(2sqrt(7))) + C.Simplify Constant: Simplify the constant in front of the arctan. 1/(4*2sqrt(7)) simplifies to 1/(8sqrt(7)). However, this is a mistake because we should not simplify the constant yet as it is part of the standard arctan integral form.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Find
int(1)/(4x^(2)+48 x+148)dx.
Choose 1 answer:
(A)
(1)/(4)arcsin(x+6)+C
(B)
(1)/(4)arcsin((x+6)/(4))+C
(C)
(1)/(4)arctan(x+6)+C
(D)
(1)/(4)arctan((x+6)/(4))+C

Find $\int \frac{1}{4 x^{2}+48 x+148} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{4} \arcsin (x+6)+C$ $\newline$ (B) $\frac{1}{4} \arcsin \left(\frac{x+6}{4}\right)+C$ $\newline$ (C) $\frac{1}{4} \arctan (x+6)+C$ $\newline$ (D) $\frac{1}{4} \arctan \left(\frac{x+6}{4}\right)+C$

View step-by-step help

Home

Math Problems

Calculus

Find derivatives of using multiple formulae

Full solution

Q. Find

\int \frac{1}{4 x^{2}+48 x+148} d x

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{4} \arcsin (x+6)+C

\newline

(B)

\frac{1}{4} \arcsin \left(\frac{x+6}{4}\right)+C

\newline

(C)

\frac{1}{4} \arctan (x+6)+C

\newline

(D)

\frac{1}{4} \arctan \left(\frac{x+6}{4}\right)+C

Complete the Square: First, complete the square for the quadratic in the denominator. $\newline$ $4x^2 + 48x + 148$ can be written as $(2x)^2 + 2\cdot2x\cdot6 + 6^2 + (148 - 36)$ . $\newline$ This simplifies to $(2x + 6)^2 + 112$ .
Rewrite with Completed Square: Now, rewrite the integral with the completed square. $\newline$ $\int\frac{1}{4x^2 + 48x + 148}\,dx = \int\frac{1}{(2x + 6)^2 + 112}\,dx$ .
Factor Out $4$ : Factor out the $4$ from the denominator to match the form of the arctan integral. $\newline$ $\int \frac{1}{(2x + 6)^2 + 112}\,dx = \int \frac{1}{4((x + 3)^2 + 28)}\,dx.$
Recognize Integral Form: Recognize that the integral is now in the form of $\int \frac{1}{a^2 + u^2}\,du$ , which is $\frac{1}{a} \arctan(\frac{u}{a}) + C$ . Here, $a^2 = 28$ and $u = x + 3$ , so $a = \sqrt{28} = 2\sqrt{7}$ .
Calculate Integral: Calculate the integral using the arctan formula. $\int \frac{1}{4((x + 3)^2 + 28)}dx = \frac{1}{4\cdot 2\sqrt{7}} \cdot \arctan\left(\frac{x + 3}{2\sqrt{7}}\right) + C$ .
Simplify Constant: Simplify the constant in front of the arctan. $\newline$ $\frac{1}{4\cdot 2\sqrt{7}}$ simplifies to $\frac{1}{8\sqrt{7}}$ . $\newline$ However, this is a mistake because we should not simplify the constant yet as it is part of the standard arctan integral form.