tolong bantu ka secepatnya pleaseee

Berikut ini adalah pertanyaan dari dimasputra2026 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Tolong bantu ka secepatnya pleaseee

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\green{\huge{1.}}$

$\int \left(2x^3+3x^2+x+7\right)~dx$

$=\frac{2}{3+1}x^{(3+1)}+\frac{3}{2+1}x^{(2+1)}+\frac{1}{1+1}x^{(1+1)}+\frac{7}{0+1}x^{(0+1)}+C$

$=\red{\huge{\frac{1}{2}x^4+x^3+\frac{1}{2}x^2+7x+C}}$

$\\$

$\green{\huge{2.}}$

$\int \left(x^2-1\right)^2~dx$

$=\int \left(x^4-2x^2+1\right)~dx$

$=\frac{1}{4+1}x^{(4+1)}-\frac{2}{2+1}x^{(2+1)}+\frac{1}{0+1}x^{(0+1)}+C$

$=\red{\huge{\frac{1}{5}x^5-\frac{2}{3}x^3+x+C}}$

$\\$

$\green{\huge{3.}}$

$\int \frac{2(x-6^2)^2}{x^5}~dx$

$=\int \frac{2\left(x^2-72x+1.296\right)}{x^5}~dx$

$=\int \frac{2x^2-144x+2.592}{x^5}~dx$

$=\int \left(2x^{-3}-144x^{-4}+2.592x^{-5}\right)~dx$

$=\frac{2}{-3+1}x^{(-3+1)}-\frac{144}{-4+1}x^{(-4+1)}+\frac{2.592}{-5+1}x^{(-5+1)}+C$

$=-x^{-2}+48x^{-3}-648x^{-4}+C$

$=\red{\huge{-\frac{1}{x^2}+\frac{48}{x^3}-\frac{648}{x^4}+C}}$

$=\red{\huge{\frac{-x^2+48x-648}{x^4}+C}}$

$\\$

$\green{\huge{4.}}$

$\int (3x+2)(x-1)~dx$

$=\int \left(3x^2-x-2\right)~dx$

$=\frac{3}{2+1}x^{(2+1)}-\frac{1}{1+1}x^{(1+1)}-\frac{2}{0+1}x^{(0+1)}+C$

$=\red{\huge{x^3-\frac{1}{2}x^2-2x+C}}$

$\\$

$\green{\huge{5.}}$

$\int \left(4x^2\sqrt{x}-3x\sqrt{x}+8\sqrt{x}\right)~dx$

$=\int \left(4x^\frac{5}{2}-3x^\frac{3}{2}+8x^\frac{1}{2}\right)~dx$

$=\frac{4}{\frac{5}{2}+1}x^{\left(\frac{5}{2}+1\right)}-\frac{3}{\frac{3}{2}+1}x^{\left(\frac{3}{2}+1\right)}+\frac{8}{\frac{1}{2}+1}x^{\left(\frac{1}{2}+1\right)}+C$

$=\frac{8}{7}x^\frac{7}{2}-\frac{6}{5}x^\frac{5}{2}+\frac{16}{3}x^\frac{3}{2}+C$

$=\red{\huge{\frac{8}{7}x^3\sqrt{x}-\frac{6}{5}x^2\sqrt{x}+\frac{16}{3}x\sqrt{x}+C}}$

$\\$

$\green{\huge{6.}}$

$\int 2\sqrt{x}~dx=\int 2x^\frac{1}{2}~dx$

$=\frac{2}{\frac{1}{2}+1}x^{\left(\frac{1}{2}+1\right)}+C=\frac{4}{3}x^\frac{3}{2}+C$

$=\red{\huge{\frac{4}{3}x\sqrt{x}+C}}$

$\\$

$\green{\huge{7.}}$

Dimisalkan : $u=2x+1$

$\frac{du}{dx}=2~\to dx=\frac{1}{2}~du$

Maka :

$\int (2x+1)^4~dx=\int u^4~\frac{1}{2}~du$

$=\frac{1}{2}\times \frac{1}{4+1}u^{(4+1)}+C$

$=\frac{1}{10}u^5+C$

$=\red{\huge{\frac{1}{10}(2x+1)^5+C}}$

$\\$

$\green{\huge{8.}}$

Dimisalkan : $u=4x^3-7$

$\frac{du}{dx}=12x^2~\to dx=\frac{1}{12x^2}~du$

Maka :

$\int \frac{48x^2}{\sqrt{4x^3-7}}~dx$

$=\int \frac{48x^2}{u^\frac{1}{2}}~\frac{1}{12x^2}~du$

$=\int 4u^{-\frac{1}{2}}~du$

$=\frac{4}{-\frac{1}{2}+1}u^{\left(-\frac{1}{2}+1\right)}+C$

$=8u^\frac{1}{2}+C$

$=\red{\huge{8\sqrt{4x^3-7}+C}}$

$\\$

$\green{\huge{9.}}$

$\begin{array}{lll}f(x)&=&\int f'(x)~dx\\~\\&=&\int \left(3x^2-5x+1\right)~dx\\~\\&=&\frac{3}{2+1}x^{(2+1)}-\frac{5}{1+1}x^{(1+1)}+\frac{1}{0+1}x^{(0+1)}+C\\~\\&=&x^3-\frac{5}{2}x^2+x+C\end{array}$

Diketahui : $f(1)=\frac{7}{2}$

$\begin{array}{rcl}1^3-\frac{5}{2}.\left(1^2\right)+1+C&=&\frac{7}{2}\\~\\1-\frac{5}{2}+1+C&=&\frac{7}{2}\\~\\\frac{1}{2}+C&=&\frac{7}{2}\\~\\C&=&3\end{array}$

Sehingga :

$\red{\huge{f(x)=x^3-\frac{5}{2}x^2+x+3}}$

$\\$

$\green{\huge{10.}}$

$\begin{array}{lll}f'(x)&=&\int f"(x)~dx\\~\\&=&\int (6x+8)~dx\\~\\&=&\frac{6}{1+1}x^{(1+1)}+\frac{8}{0+1}x^{(0+1)}+C_1\\~\\&=&3x^2+4x+C_1\end{array}$

$\begin{array}{rcl}f'(3)&=&7\\~\\3.\left(3^2\right)+4.(3)+C_1&=&7\\~\\27+12+C_1&=&7\\~\\C_1&=&-32\end{array}$

$f'(x)=3x^2+4x-32$

$\begin{array}{lll}f(x)&=&\int f'(x)~dx\\~\\&=&\int \left(3x^2+4x-32\right)~dx\\~\\&=&\frac{3}{2+1}x^{(2+1)}+\frac{4}{1+1}x^{(1+1)}-\frac{32}{0+1}x^{(0+1)}+C_2\\~\\&=&x^3+2x^2-32x+C_2\end{array}$