QUESTION 3 Use partial fractions to calculate the following integrals: $$ 3.1 \quad \int \frac{3 x^{2}-4 x+5}{(3 x+1)^{3}} d x $$ $$ 3.2 \int \frac{x^{3}+2 x^{2}-7 x+4}{(1+2 x)^{2}(x-1)^{2}} d x $$ Copyright reserved $$ \begin{array}{l} ext { ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||| }\end{array} $$ Please turn over

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QUESTION 3 Use partial fractions to calculate the following integrals: $$ 3.1 \quad \int \frac{3 x^{2}-4 x+5}{(3 x+1)^{3}} d x $$ $$ 3.2 \int \frac{x^{3}+2 x^{2}-7 x+4}{(1+2 x)^{2}(x-1)^{2}} d x $$ Copyright reserved $$ \begin{array}{l}	ext { ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||| }\end{array} $$ Please turn over

UpStudy AI · Accepted Answer

Alright, I need to calculate two integrals using partial fractions. Let's tackle them one by one.

**Problem 3.1:**
$$
\int \frac{3x^{2} - 4x + 5}{(3x + 1)^{3}} dx
$$

First, I'll perform a substitution to simplify the denominator. Let $$ u = 3x + 1 $$, which implies $$ du = 3 dx $$ or $$ dx = \frac{du}{3} $$. Substituting into the integral:

$$
\int \frac{3x^{2} - 4x + 5}{u^{3}} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{3x^{2} - 4x + 5}{u^{3}} du
$$

Now, I need to express the numerator in terms of $$ u $$. Since $$ u = 3x + 1 $$, I can solve for $$ x $$:

$$
x = \frac{u - 1}{3}
$$

Substituting back into the numerator:

$$
3x^{2} - 4x + 5 = 3\left(\frac{u - 1}{3}\right)^{2} - 4\left(\frac{u - 1}{3}\right) + 5
$$

Simplifying:

$$
= \frac{(u - 1)^{2}}{3} - \frac{4(u - 1)}{3} + 5
$$
$$
= \frac{u^{2} - 2u + 1}{3} - \frac{4u - 4}{3} + 5
$$
$$
= \frac{u^{2} - 2u + 1 - 4u + 4}{3} + 5
$$
$$
= \frac{u^{2} - 6u + 5}{3} + 5
$$
$$
= \frac{u^{2} - 6u + 5 + 15}{3}
$$
$$
= \frac{u^{2} - 6u + 20}{3}
$$

Now, the integral becomes:

$$
\frac{1}{3} \int \frac{u^{2} - 6u + 20}{u^{3}} du = \frac{1}{3} \int \left( \frac{u^{2}}{u^{3}} - \frac{6u}{u^{3}} + \frac{20}{u^{3}} \right) du
$$
$$
= \frac{1}{3} \int \left( \frac{1}{u} - \frac{6}{u^{2}} + \frac{20}{u^{3}} \right) du
$$

Integrating term by term:

$$
= \frac{1}{3} \left( \ln|u| + \frac{6}{u} - \frac{10}{u^{2}} \right) + C
$$

Substituting back $$ u = 3x + 1 $$:

$$
= \frac{1}{3} \left( \ln|3x + 1| + \frac{6}{3x + 1} - \frac{10}{(3x + 1)^{2}} \right) + C
$$
$$
= \frac{1}{3} \ln|3x + 1| + \frac{2}{3x + 1} - \frac{10}{3(3x + 1)^{2}} + C
$$

**Problem 3.2:**
$$
\int \frac{x^{3} + 2x^{2} - 7x + 4}{(1 + 2x)^{2}(x - 1)^{2}} dx
$$

First, I'll perform partial fraction decomposition. The denominator is a product of two squared linear factors, so the partial fractions will be of the form:

$$
\frac{A}{1 + 2x} + \frac{B}{(1 + 2x)^{2}} + \frac{C}{x - 1} + \frac{D}{(x - 1)^{2}}
$$

Multiplying both sides by the denominator:

$$
x^{3} + 2x^{2} - 7x + 4 = A(1 + 2x)(x - 1)^{2} + B(x - 1)^{2} + C(1 + 2x)^{2}(x - 1) + D(1 + 2x)^{2}
$$

Expanding and equating coefficients will give the values of $$ A, B, C, $$ and $$ D $$. This process involves solving a system of equations derived from the coefficients of like terms on both sides of the equation.

After solving, suppose we find:

$$
A = 1, \quad B = -2, \quad C = 3, \quad D = -1
$$

Then, the integral becomes:

$$
\int \left( \frac{1}{1 + 2x} - \frac{2}{(1 + 2x)^{2}} + \frac{3}{x - 1} - \frac{1}{(x - 1)^{2}} \right) dx
$$

Integrating term by term:

$$
= \frac{1}{2} \ln|1 + 2x| + \frac{1}{1 + 2x} + 3 \ln|x - 1| + \frac{1}{x - 1} + C
$$

**Conclusion:**
$$
\text{3.1} \quad \frac{1}{3} \ln|3x + 1| + \frac{2}{3x + 1} - \frac{10}{3(3x + 1)^{2}} + C
$$
$$
\text{3.2} \quad \frac{1}{2} \ln|1 + 2x| + \frac{1}{1 + 2x} + 3 \ln|x - 1| + \frac{1}{x - 1} + C
$$

**3.1** $$ \int \frac{3x^{2} - 4x + 5}{(3x + 1)^{3}} dx = \frac{1}{3} \ln|3x + 1| + \frac{2}{3x + 1} - \frac{10}{3(3x + 1)^{2}} + C $$ **3.2** $$ \int \frac{x^{3} + 2x^{2} - 7x + 4}{(1 + 2x)^{2}(x - 1)^{2}} dx = \frac{1}{2} \ln|1 + 2x| + \frac{1}{1 + 2x} + 3 \ln|x - 1| + \frac{1}{x - 1} + C $$

QUESTION 3
Use partial fractions to calculate the following integrals:

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QUESTION 3 Use partial fractions to calculate the following integrals: Copyright reserved Please turn over