Integration: Learning Objectives 7, 8

The integrand is a simple polynomial. You can integrate term-by-term using the Power Rule, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$.

$37$

Apply the power rule for integration to each term: $$ \int_1^2 (8x^3 + 3x^2) dx = \left[ 8\frac{x^4}{4} + 3\frac{x^3}{3} \right]_1^2 $$ $$ = [2x^4 + x^3]_1^2 $$ Now, evaluate using the Fundamental Theorem of Calculus: $$ = (2(2)^4 + (2)^3) - (2(1)^4 + (1)^3) $$ $$ = (2(16) + 8) - (2(1) + 1) $$ $$ = (32 + 8) - (2 + 1) = 40 - 3 = 37 $$

The integrand is a product of a polynomial and an exponential function. This is a classic case for Integration by Parts, $\int u \, dv = uv - \int v \, du$. A good choice for $u$ is the polynomial part.

$-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C$

We use integration by parts. Let:

• $u = x$
• $dv = e^{-2x} dx$

Then we find:

• $du = dx$
• $v = \int e^{-2x} dx = -\frac{1}{2}e^{-2x}$

Apply the formula $\int u \, dv = uv - \int v \, du$: $$ \int xe^{-2x} dx = x \left(-\frac{1}{2}e^{-2x}\right) - \int \left(-\frac{1}{2}e^{-2x}\right) dx $$ $$ = -\frac{1}{2}xe^{-2x} + \frac{1}{2} \int e^{-2x} dx $$ $$ = -\frac{1}{2}xe^{-2x} + \frac{1}{2} \left(-\frac{1}{2}e^{-2x}\right) + C $$ $$ = -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C $$

Notice that the derivative of the inner function, $y^2+1$, is $2y$, which is a constant multiple of the factor $y$ outside the parenthesis. This suggests using u-substitution.

$\frac{21}{4}$

Let $u = y^2 + 1$. Then $du = 2y \, dy$, which means $y \, dy = \frac{1}{2} du$.

We must also change the limits of integration:

• When $y = 0$, $u = (0)^2 + 1 = 1$.
• When $y = 1$, $u = (1)^2 + 1 = 2$.

Now substitute into the integral: $$ \int_0^1 y(y^2 + 1)^5 dy = \int_1^2 u^5 \left(\frac{1}{2} du\right) = \frac{1}{2} \int_1^2 u^5 du $$ $$ = \frac{1}{2} \left[ \frac{u^6}{6} \right]_1^2 = \frac{1}{12} [u^6]_1^2 $$ $$ = \frac{1}{12} (2^6 - 1^6) = \frac{1}{12} (64 - 1) = \frac{63}{12} = \frac{21}{4} $$

This requires a multi-step approach. Start with a u-substitution to handle the root in the exponent. Let $u = \sqrt[3]{x}$. After substituting, the new integral will require integration by parts, potentially more than once.

$3e^{\sqrt[3]{x}}(x^{2/3} - 2x^{1/3} + 2) + C$

First, let $u = \sqrt[3]{x}$. This means $x = u^3$, so $dx = 3u^2 du$. Substituting gives: $$ \int e^{\sqrt[3]{x}} dx = \int e^u (3u^2 du) = 3 \int u^2 e^u du $$ This new integral requires integration by parts. Let $U = u^2$ and $dV = e^u du$. Then $dU = 2u du$ and $V = e^u$. $$ 3 \left[ u^2e^u - \int 2u e^u du \right] = 3u^2e^u - 6 \int u e^u du $$ We must apply integration by parts again to the remaining integral. Let $U_2 = u$ and $dV_2 = e^u du$. Then $dU_2 = du$ and $V_2 = e^u$. $$ \int u e^u du = ue^u - \int e^u du = ue^u - e^u $$ Substitute this back into our expression: $$ 3u^2e^u - 6(ue^u - e^u) + C = 3u^2e^u - 6ue^u + 6e^u + C $$ Factor out $3e^u$: $$ = 3e^u(u^2 - 2u + 2) + C $$ Finally, substitute back $u = \sqrt[3]{x} = x^{1/3}$: $$ = 3e^{x^{1/3}}((x^{1/3})^2 - 2x^{1/3} + 2) + C = 3e^{\sqrt[3]{x}}(x^{2/3} - 2x^{1/3} + 2) + C $$

The integrand is a rational function where the denominator can be factored into distinct linear terms. This is a prime candidate for the method of Partial Fraction Decomposition.

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

First, factor the denominator: $x^2+2x = x(x+2)$. We can then decompose the fraction: $$ \frac{x-1}{x(x+2)} = \frac{A}{x} + \frac{B}{x+2} $$ To find the constants $A$ and $B$, multiply both sides by $x(x+2)$: $$ x-1 = A(x+2) + Bx $$ We can solve for $A$ and $B$ by picking convenient values for $x$:

• Let $x=0$: $0-1 = A(0+2) + B(0) \implies -1 = 2A \implies A = -1/2$.
• Let $x=-2$: $-2-1 = A(-2+2) + B(-2) \implies -3 = -2B \implies B = 3/2$.

Now, substitute the decomposed form back into the integral: $$ \int \left(-\frac{1}{2x} + \frac{3}{2(x+2)}\right) dx = -\frac{1}{2}\int \frac{1}{x} dx + \frac{3}{2}\int \frac{1}{x+2} dx $$ $$ = -\frac{1}{2}\ln|x| + \frac{3}{2}\ln|x+2| + C $$

Before integrating, perform algebraic simplification. Divide each term in the numerator by the denominator, $u$, to get a simpler expression that can be integrated with the power rule.

$-76$

First, simplify the integrand: $$ \frac{\sqrt{u}-2u^2}{u} = \frac{u^{1/2}}{u} - \frac{2u^2}{u} = u^{-1/2} - 2u $$ Now integrate this simplified expression using the power rule: $$ \int_1^9 (u^{-1/2} - 2u) du = \left[ \frac{u^{1/2}}{1/2} - 2\frac{u^2}{2} \right]_1^9 $$ $$ = [2\sqrt{u} - u^2]_1^9 $$ Evaluate at the limits: $$ = (2\sqrt{9} - 9^2) - (2\sqrt{1} - 1^2) $$ $$ = (2(3) - 81) - (2 - 1) = (6 - 81) - 1 = -75 - 1 = -76 $$

This problem is best solved with Integration by Parts. Try setting $dv = \frac{1}{(1+2x)^2}dx$. The key is that the derivative of $u$ will simplify nicely with the resulting $v$.

$\frac{e-2}{8}$

We use integration by parts. Let:

• $u = xe^{2x}$
• $dv = (1+2x)^{-2} dx$

Then we find (using the product rule for $du$):

• $du = (1 \cdot e^{2x} + x \cdot 2e^{2x}) dx = (1+2x)e^{2x} dx$
• $v = \int (1+2x)^{-2} dx = -\frac{1}{2}(1+2x)^{-1} = -\frac{1}{2(1+2x)}$

Apply the IBP formula $\int u \, dv = uv - \int v \, du$: $$ \int_0^{1/2} \frac{xe^{2x}}{(1+2x)^2} dx = \left[ -\frac{xe^{2x}}{2(1+2x)} \right]_0^{1/2} - \int_0^{1/2} \left(-\frac{1}{2(1+2x)}\right) (1+2x)e^{2x} dx $$ The second integral simplifies beautifully: $$ = \left[ -\frac{xe^{2x}}{2(1+2x)} \right]_0^{1/2} + \frac{1}{2} \int_0^{1/2} e^{2x} dx $$ Now, evaluate both parts: $$ = \left( -\frac{(1/2)e^{2(1/2)}}{2(1+2(1/2))} - 0 \right) + \frac{1}{2} \left[ \frac{1}{2}e^{2x} \right]_0^{1/2} $$ $$ = \left( -\frac{e/2}{2(2)} \right) + \frac{1}{4} [e^{2x}]_0^{1/2} = -\frac{e}{8} + \frac{1}{4}(e^1 - e^0) $$ $$ = -\frac{e}{8} + \frac{1}{4}(e - 1) = \frac{-e + 2(e-1)}{8} = \frac{-e+2e-2}{8} = \frac{e-2}{8} $$

This is an improper integral due to the infinite upper limit. You must first set it up as a limit. The resulting integral can be solved with a u-substitution where $u = \ln(x)$.

The integral diverges.

First, we express the improper integral as a limit: $$ \int_2^\infty \frac{1}{x\ln(x)} dx = \lim_{t \to \infty} \int_2^t \frac{1}{x\ln(x)} dx $$ To solve the integral, we use the substitution $u = \ln(x)$, which gives $du = \frac{1}{x} dx$. $$ \int \frac{1}{x\ln(x)} dx = \int \frac{1}{u} du = \ln|u| = \ln|\ln(x)| $$ Now, we evaluate the definite integral within the limit: $$ \lim_{t \to \infty} \left[ \ln|\ln(x)| \right]_2^t = \lim_{t \to \infty} (\ln(\ln(t)) - \ln(\ln(2))) $$ As $t \to \infty$, $\ln(t) \to \infty$, and therefore $\ln(\ln(t)) \to \infty$. Since the limit is infinite, the integral diverges.

The most direct approach is algebraic simplification. Expand the numerator $(x+1)^2$ first, then divide each resulting term by the denominator $x$.

$\frac{7}{2} + \ln(2)$

First, expand the numerator: $(x+1)^2 = x^2 + 2x + 1$.

Now, divide the entire expression by $x$: $$ \frac{x^2 + 2x + 1}{x} = \frac{x^2}{x} + \frac{2x}{x} + \frac{1}{x} = x + 2 + \frac{1}{x} $$ Integrate the simplified expression: $$ \int_1^2 \left(x + 2 + \frac{1}{x}\right) dx = \left[ \frac{x^2}{2} + 2x + \ln|x| \right]_1^2 $$ Evaluate at the limits: $$ = \left( \frac{2^2}{2} + 2(2) + \ln(2) \right) - \left( \frac{1^2}{2} + 2(1) + \ln(1) \right) $$ $$ = (2 + 4 + \ln(2)) - (\frac{1}{2} + 2 + 0) = (6 + \ln(2)) - 2.5 $$ $$ = 3.5 + \ln(2) = \frac{7}{2} + \ln(2) $$

This integral is best tackled with u-substitution. Let $u = 1 + \sqrt{x}$. You will need to solve for $x$ and $dx$ in terms of $u$ and $du$.

$\frac{4097}{45}$

Let $u = 1 + \sqrt{x}$. Then $\sqrt{x} = u - 1$, so $x = (u-1)^2$.
Differentiating to find $dx$: $dx = 2(u-1) du$.

Now, change the limits of integration:

• When $x = 0$, $u = 1 + \sqrt{0} = 1$.
• When $x = 1$, $u = 1 + \sqrt{1} = 2$.

Substitute into the integral: $$ \int_0^1 (1 + \sqrt{x})^8 dx = \int_1^2 u^8 \cdot 2(u-1) du = 2 \int_1^2 (u^9 - u^8) du $$ Now integrate: $$ = 2 \left[ \frac{u^{10}}{10} - \frac{u^9}{9} \right]_1^2 $$ Evaluate at the limits: $$ = 2 \left[ \left(\frac{2^{10}}{10} - \frac{2^9}{9}\right) - \left(\frac{1^{10}}{10} - \frac{1^9}{9}\right) \right] $$ $$ = 2 \left[ \left(\frac{1024}{10} - \frac{512}{9}\right) - \left(\frac{1}{10} - \frac{1}{9}\right) \right] $$ $$ = 2 \left[ \frac{1023}{10} - \frac{511}{9} \right] = 2 \left[ \frac{1023 \cdot 9 - 511 \cdot 10}{90} \right] $$ $$ = 2 \left[ \frac{9207 - 5110}{90} \right] = 2 \left[ \frac{4097}{90} \right] = \frac{4097}{45} $$