Q.$\int_{0}^{1} (x + 2x + 3x^2 + 4x^3)\,dx =$

Integrate each term: $\int(x+2)\,dx=\tfrac{x^2}{2}+2x+k$.

Apply the power rule $\int x^n\,dx=\tfrac{x^{n+1}}{n+1}$ to each term:

…

Integrate term by term using standard integrals.

$\displaystyle\int(2x-3\cos x+e^x)\,dx = \int2x\,dx - 3\int\cos x\,dx+\int e^x\,dx$

…

Combine $x+2x=3x$; integrate term by term over $[0,1]$: $\frac32+1+1=\frac72$.

First simplify: $x + 2x + 3x^2 + 4x^3 = 3x + 3x^2 + 4x^3.$ Integrate:

…

Expand and integrate the standard forms to get $\tan x+\sec x+c$, so (b). …

$\int(x+\cos 2x)\,dx=\frac{x^2}{2}+\frac{\sin 2x}{2}+c$.

Integrate each term:

$\int x\,dx = \dfrac{x^2}{2}$, and $\int \cos 2x\,dx = \dfrac{\sin 2x}{2}$.

…

Integrate each term separately using the standard formulas $\int x\,dx=\dfrac{x^2}{2}$ scaled by 2, $\int\cos x\,dx=\sin x$, and $\int e^x\,dx=e^x$.

$\displaystyle\int (2x - 3\cos x + e^x)\,dx = 2\int x\,dx - 3\int \cos x\,dx + \int e^x\,dx$

…

$\int4\cos x\,dx=4\sin x$ and $\int(-5\sin x)\,dx=+5\cos x$, summing to $4\sin x+5\cos x+k$.

Integrate each term:

$\int 4\cos x\,dx=4\sin x$.

$\int (-5\sin x)\,dx=-5(-\cos x)=5\cos x$.

…

Expand the integrand into standard forms $\sec^2x$ and $\sec x\tan x$. …

Integrate $x$ and $2$ separately.

…

$\displaystyle\int_0^2 (x^2+1)\,dx = \dfrac{14}{3}$.

Integrate term by term:

$\int_0^2 (x^2+1)\,dx = \left[\dfrac{x^3}{3} + x\right]_0^2.$

Evaluate at the limits: …

$\displaystyle\int(2x^2+e^x)\,dx=\dfrac{2x^3}{3}+e^x+C$.

Concept. Integration is linear, so we may integrate each term separately and pull out constants.

Step-by-step.

$\int 2x^2\,dx=2\cdot\frac{x^{3}}{3}=\frac{2x^3}{3},\qquad \int e^x\,dx=e^x.$ …