Integration: Learning Objective 5

Let $u$ be the expression inside the square root, $u = 1+x^2$. Notice that its derivative, $du = 2x\,dx$, is present as a factor in the integrand.

\( \frac{2}{3}(1+x^2)^{3/2} + C \)

Let $u = 1+x^2$. Then, $du = 2x\,dx$. Substituting these into the integral gives: \[ \int \sqrt{1 + x^2} (2x \, dx) = \int \sqrt{u} \, du \] Now, integrate with respect to $u$ using the power rule: \[ = \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} + C = \frac{2}{3}u^{3/2} + C \] Finally, substitute back for $u$: \[ = \frac{2}{3}(1+x^2)^{3/2} + C \]

Let $u$ be the argument of the cosine function, $u = x^4+2$. Its derivative is nearly present in the integrand.

\( \frac{1}{4}\sin(x^4+2) + C \)

Let $u = x^4+2$. Then, $du = 4x^3\,dx$. This means that $x^3\,dx = \frac{1}{4}du$. Substituting these gives: \[ \int \cos(x^4+2) (x^3 \, dx) = \int \cos(u) \left(\frac{1}{4}du\right) = \frac{1}{4}\int \cos(u) \, du \] Integrate with respect to $u$: \[ = \frac{1}{4}\sin(u) + C \] Substitute back for $u$: \[ = \frac{1}{4}\sin(x^4+2) + C \]

Let $u$ be the expression inside the square root, $u=1-4x^2$.

\( -\frac{1}{4}\sqrt{1-4x^2} + C \)

Let $u = 1-4x^2$. Then, $du = -8x\,dx$, which means $x\,dx = -\frac{1}{8}du$. \[ \int \frac{1}{\sqrt{1-4x^2}} (x \, dx) = \int \frac{1}{\sqrt{u}} \left(-\frac{1}{8}du\right) = -\frac{1}{8}\int u^{-1/2} \, du \] Integrate with respect to $u$: \[ = -\frac{1}{8} \left(\frac{u^{1/2}}{1/2}\right) + C = -\frac{1}{8}(2\sqrt{u}) + C = -\frac{1}{4}\sqrt{u} + C \] Substitute back for $u$: \[ = -\frac{1}{4}\sqrt{1-4x^2} + C \]

A simple substitution is needed here. Let $u$ be the exponent, $u=5x$.

\( \frac{1}{5}e^{5x} + C \)

Let $u=5x$. Then $du = 5\,dx$, which means $dx = \frac{1}{5}du$. \[ \int e^{5x} \, dx = \int e^u \left(\frac{1}{5}du\right) = \frac{1}{5}\int e^u \, du \] Integrate with respect to $u$: \[ = \frac{1}{5}e^u + C \] Substitute back for $u$: \[ = \frac{1}{5}e^{5x} + C \]

Let $u$ be the exponent of $e$, so $u = 1/x$.

\( -e^{1/x} + C \)

Let $u = \frac{1}{x} = x^{-1}$. Then $du = -x^{-2}dx = -\frac{1}{x^2}dx$. This means $\frac{1}{x^2}dx = -du$. \[ \int e^{1/x} \left(\frac{1}{x^2} \, dx\right) = \int e^u (-du) = -\int e^u \, du \] Integrate with respect to $u$: \[ = -e^u + C \] Substitute back for $u$: \[ = -e^{1/x} + C \]

Let $u=1+x^2$. You will need to split $x^5$ into $x^4 \cdot x = (x^2)^2 \cdot x$ and express $x^2$ in terms of $u$.

\( \frac{1}{7}(1+x^2)^{7/2} - \frac{2}{5}(1+x^2)^{5/2} + \frac{1}{3}(1+x^2)^{3/2} + C \)

Let $u = 1+x^2$. This implies $x^2 = u-1$ and $du = 2x\,dx$, so $x\,dx = \frac{1}{2}du$. We rewrite the integrand by splitting $x^5$: \[ \int \sqrt{1+x^2}x^5 \, dx = \int \sqrt{1+x^2} \cdot x^4 \cdot x \, dx = \int \sqrt{1+x^2} (x^2)^2 (x \, dx) \] Now substitute: \[ = \int \sqrt{u}(u-1)^2 \left(\frac{1}{2}du\right) = \frac{1}{2}\int u^{1/2}(u^2-2u+1) \, du \] Distribute the $u^{1/2}$: \[ = \frac{1}{2}\int (u^{5/2} - 2u^{3/2} + u^{1/2}) \, du \] Integrate term by term: \[ = \frac{1}{2}\left(\frac{u^{7/2}}{7/2} - 2\frac{u^{5/2}}{5/2} + \frac{u^{3/2}}{3/2}\right) + C = \frac{1}{2}\left(\frac{2}{7}u^{7/2} - \frac{4}{5}u^{5/2} + \frac{2}{3}u^{3/2}\right) + C \] Simplify and substitute back for $u$: \[ = \frac{1}{7}(1+x^2)^{7/2} - \frac{2}{5}(1+x^2)^{5/2} + \frac{1}{3}(1+x^2)^{3/2} + C \]

Let $u = \ln(x)$.

\( 2\sqrt{\ln(x)} + C \)

Let $u = \ln(x)$. Then $du = \frac{1}{x}dx$. \[ \int \frac{1}{\sqrt{\ln(x)}} \left(\frac{1}{x}dx\right) = \int \frac{1}{\sqrt{u}} \, du = \int u^{-1/2} \, du \] Integrate: \[ = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C \] Substitute back: \[ = 2\sqrt{\ln(x)} + C \]

Rewrite $\tan(x)$ as $\frac{\sin(x)}{\cos(x)}$ and then use a substitution with $u = \cos(x)$.

\( \ln|\sec(x)| + C \) or \( -\ln|\cos(x)| + C \)

First, rewrite the integral: \[ \int \tan(x) \, dx = \int \frac{\sin(x)}{\cos(x)} \, dx \] Let $u = \cos(x)$. Then $du = -\sin(x)dx$, which means $\sin(x)dx = -du$. \[ = \int \frac{1}{u} (-du) = -\int \frac{1}{u} \, du \] Integrate: \[ = -\ln|u| + C \] Substitute back for $u$: \[ = -\ln|\cos(x)| + C \] Using logarithm properties, this can also be written as: \[ = \ln(|\cos(x)|^{-1}) + C = \ln\left|\frac{1}{\cos(x)}\right| + C = \ln|\sec(x)| + C \]

Let $u = a^2 - x^2$.

\( -\frac{1}{3}(a^2 - x^2)^{3/2} + C \)

Let $u = a^2 - x^2$. Then $du = -2x\,dx$, which means $x\,dx = -\frac{1}{2}du$. \[ \int \sqrt{a^2-x^2}(x\,dx) = \int \sqrt{u}\left(-\frac{1}{2}du\right) = -\frac{1}{2}\int u^{1/2}\,du \] Integrate: \[ = -\frac{1}{2}\left(\frac{u^{3/2}}{3/2}\right) + C = -\frac{1}{2}\left(\frac{2}{3}u^{3/2}\right) + C = -\frac{1}{3}u^{3/2} + C \] Substitute back: \[ = -\frac{1}{3}(a^2 - x^2)^{3/2} + C \]

This is an improper integral. First, find the indefinite integral using the substitution $u = 1+x$. Then, evaluate the definite integral using limits. Remember to change the bounds of integration when you substitute.

\( \frac{1}{2} \)

We first set up the limit for the improper integral: \[ \lim_{t \to \infty} \int_{0}^{t} \frac{3x}{(1+x)^4} \, dx \] Now we perform the substitution. Let $u=1+x$. Then $du=dx$ and $x=u-1$. We must also change the limits of integration:

• When $x=0$, $u=1+0=1$.
• When $x=t$, $u=1+t$.

The integral becomes: \[ \lim_{t \to \infty} \int_{1}^{1+t} \frac{3(u-1)}{u^4} \, du = 3 \lim_{t \to \infty} \int_{1}^{1+t} \left(\frac{u}{u^4} - \frac{1}{u^4}\right) \, du = 3 \lim_{t \to \infty} \int_{1}^{1+t} (u^{-3} - u^{-4}) \, du \] Integrate: \[ = 3 \lim_{t \to \infty} \left[ \frac{u^{-2}}{-2} - \frac{u^{-3}}{-3} \right]_{1}^{1+t} = 3 \lim_{t \to \infty} \left[ -\frac{1}{2u^2} + \frac{1}{3u^3} \right]_{1}^{1+t} \] Apply the limits of integration: \[ = 3 \lim_{t \to \infty} \left( \left(-\frac{1}{2(1+t)^2} + \frac{1}{3(1+t)^3}\right) - \left(-\frac{1}{2(1)^2} + \frac{1}{3(1)^3}\right) \right) \] As $t \to \infty$, the terms with $t$ in the denominator go to 0: \[ = 3 \left( (0 + 0) - (-\frac{1}{2} + \frac{1}{3}) \right) = 3 \left( -(-\frac{3}{6} + \frac{2}{6}) \right) = 3 \left( -(-\frac{1}{6}) \right) = 3\left(\frac{1}{6}\right) = \frac{1}{2} \]