Integration by Substitution

$u = x^4 + 2$.

$du = 4x^3\,dx$, so $x^3\,dx = \tfrac{1}{4}\,du$.

$\displaystyle\int \tfrac{1}{4}u^5\,du$.

$\dfrac{u^6}{24}$.

$\dfrac{(x^4 + 2)^6}{24} + C$.

$u = \log x$.

$du = \dfrac{1}{x}\,dx$: exactly what is there.

$\displaystyle\int u^3\,du$.

$\dfrac{u^4}{4}$.

$\dfrac{(\log x)^4}{4} + C$.

$u = 1 + e^{2x}$, the whole denominator.

$du = 2e^{2x}\,dx$, so $e^{2x}\,dx = \tfrac{1}{2}\,du$.

$\dfrac{1}{2}\displaystyle\int \frac{du}{u}$.

$\dfrac{1}{2}\log|u|$.

$\dfrac{1}{2}\log\left(1 + e^{2x}\right) + C$. No absolute value is needed: $1 + e^{2x} > 0$.

$u = \cos x$.

$du = -\sin x\,dx$, so $\sin x\,dx = -du$.

$-\displaystyle\int u^4\,du$.

$-\dfrac{u^5}{5}$.

$-\dfrac{\cos^5 x}{5} + C$.

$u = x^3 + 1$, $\ du = 3x^2\,dx$.

$x = 0 \Rightarrow u = 1$; $\ x = 1 \Rightarrow u = 2$.

$\dfrac{1}{3}\displaystyle\int_1^2 \frac{du}{u}$.

$\dfrac{1}{3}\left[\log u\right]_1^2 = \dfrac{1}{3}\log 2$.

$u = 1 + \log x$, $\ du = \dfrac{1}{x}\,dx$.

$x = 1 \Rightarrow u = 1$; $\ x = e \Rightarrow u = 2$.

$\displaystyle\int_1^2 u^2\,du$.

$\left[\dfrac{u^3}{3}\right]_1^2 = \dfrac{8 - 1}{3} = \dfrac{7}{3}$.

$u = \cos x$, $\ du = -\sin x\,dx$.

$x = 0 \Rightarrow u = 1$; $\ x = \tfrac{\pi}{3} \Rightarrow u = \tfrac{1}{2}$.

$-\displaystyle\int_1^{1/2} u^{-2}\,du = \int_{1/2}^{1} u^{-2}\,du$.

$\left[-\dfrac{1}{u}\right]_{1/2}^{1} = -1 + 2 = 1$.

Nothing to rewrite.

$u = x + 1$, $\ du = dx$.

$x = u - 1$: $\ \displaystyle\int (u - 1)u^{1/2}\,du = \int \left(u^{3/2} - u^{1/2}\right)\,du$.

$\dfrac{2}{5}u^{5/2} - \dfrac{2}{3}u^{3/2} = \dfrac{2}{5}(x + 1)^{5/2} - \dfrac{2}{3}(x + 1)^{3/2} + C$.

$\displaystyle\int \frac{\sin x}{\cos x}\,dx$.

$u = \cos x$, $\ du = -\sin x\,dx$.

No leftover $x$: $\ -\displaystyle\int \frac{du}{u}$.

$-\log|u| = -\log|\cos x| + C = \log|\sec x| + C$.

$x^4 = (x^2)^2$, so the bottom is $1 + (x^2)^2$.

$u = x^2$, $\ du = 2x\,dx$.

$x\,dx = \tfrac{1}{2}\,du$: $\ \dfrac{1}{2}\displaystyle\int \frac{du}{1 + u^2}$.

$\dfrac{1}{2}\arctan u = \dfrac{1}{2}\arctan(x^2) + C$.

$u = x^2$.

$du = 2x\,dx$; $\ x = 0 \Rightarrow u = 0$, $\ x = 3 \Rightarrow u = 9$.

$\dfrac{1}{2}\displaystyle\int_0^9 f(u)\,du$.

$\dfrac{1}{2}\cdot 6 = 3$.

$u = \sin x$.

$du = \cos x\,dx$; $\ x = 0 \Rightarrow u = 0$, $\ x = \tfrac{\pi}{2} \Rightarrow u = 1$.

$\displaystyle\int_0^1 f(u)\,du$.

$4$.

$u = e^{x}$.

$du = e^{x}\,dx = u\,dx$; $\ x = 0 \Rightarrow u = 1$, $\ x = \log 2 \Rightarrow u = 2$.

$dx = \dfrac{du}{u}$: $\ \displaystyle\int_1^2 \frac{f(u)}{u}\,du$.

No value given: this is the answer.