Question

Решить интегралы, желательно подробно.

Answer 1

$11);int(x^2+3)^5xdx=left(begin{array}{c}u=x^2+3\du=2xdxend{array}right)=frac12int u^5du=frac12cdotfrac{u^6}6=frac{u^6}{12}=\=frac{(x^2+3)^6}{12}\12);intfrac{(x^2-7)^3x}2dx=frac12int(x^2-7)^3xdx=left(begin{array}{c}u=x^2-7\du=2xdxend{array}right)=\=frac14int u^3dx=frac14cdotfrac{u^4}4=frac{u^5}{16}=frac{(x^2-7)^5}{16}\13);intsqrt{(x^4-3)^3}x^3dx=left(begin{array}{c}u=x^4-1\du=4x^3dxend{array}right)=frac14int u^{frac32}du=frac14cdotfrac25u^{frac52}=$
$=frac{u^frac52}{20}=frac{sqrt{(x^4-1)^5}}{20}$
$14);intfrac{cos xdx}{5sin x+2}=left(begin{array}{c}u=5sin x+2\du=5cos xdxend{array}right)=frac15intfrac{du}u=frac15ln u=\=frac15ln(5sin x+2)\15);intfrac{dx}{sqrt{(3x-2)^3}}=left(begin{array}{c}u=3x-2\du=3dxend{array}right)=frac13intfrac{du}{u^{frac32}}=frac13cdotfrac{-2}{u^{frac13}}=-frac2{3u^{frac12}}=\=-frac2{3sqrt{3x-2}}\16);intsqrt{e^x+1}e^xdx=left(begin{array}{c}u=e^x+1\du=e^xdxend{array}right)=intsqrt udu=frac23 u^{frac32}=$
$=frac{2sqrt{(e^x+1)}}{3}\17);intsqrt[3]{(5-2x)^2}dx=left(begin{array}{c}u=5-2x\du=-2dxend{array}right)=-frac12int u^{frac23}du=-frac12cdotfrac35 u^{frac53}=\=-frac{3u^{frac53}}{10}=-frac{3sqrt[3]{(5-2x)^5}}{10}$